Count pairs from two BSTs whose sum is equal to a given value x

Given two BSTs containing n1 and n2 distinct nodes respectively. Given a value x. The problem is to count all pairs from both the BSTs whose sum is equal to x.

Examples: 

Input : BST 1:    5        
                /   \      
               3     7      
              / \   / \    
             2  4  6   8   

        BST 2:    10        
                /   \      
               6     15      
              / \   /  \    
             3  8  11  18
        x = 16
    
Output : 3
The pairs are:
(5, 11), (6, 10) and (8, 8)

Method 1: For each node value a in BST 1, search the value (x – a) in BST 2. If value found then increment the count. For searching a value in BST.

Implementation

C++

#include <bits/stdc++.h>
#include <iostream>
#include <unordered_set>
using namespace std;

// Definition of BST node
struct Node {
    int data;
    Node *left, *right;
    Node(int x)
    {
        data = x;
        left = right = NULL;
    }
};

// Function to insert a node in BST
Node* insert(Node* root, int x)
{
    if (root == NULL)
        return new Node(x);
    if (x < root->data)
        root->left = insert(root->left, x);
    else if (x > root->data)
        root->right = insert(root->right, x);
    return root;
}

// Function to count pairs with sum equal to x
int countPairs(Node* root1, Node* root2, int x)
{
    // Set to store values of BST 2
    unordered_set<int> s;

    // Traverse BST 2 and insert values in the set
    stack<Node*> st;
    Node* curr = root2;
    while (curr != NULL || !st.empty()) {
        while (curr != NULL) {
            st.push(curr);
            curr = curr->left;
        }
        curr = st.top();
        st.pop();
        s.insert(curr->data);
        curr = curr->right;
    }

    // Traverse BST 1 and search for (x-a) in the set
    int count = 0;
    curr = root1;
    while (curr != NULL || !st.empty()) {
        while (curr != NULL) {
            st.push(curr);
            curr = curr->left;
        }
        curr = st.top();
        st.pop();
        if (s.find(x - curr->data) != s.end())
            count++;
        curr = curr->right;
    }

    return count;
}

// Driver code
int main()
{
    // Input BSTs
    Node *root1 = NULL, *root2 = NULL;
    root1 = insert(root1, 5);
    root1 = insert(root1, 3);
    root1 = insert(root1, 7);
    root1 = insert(root1, 2);
    root1 = insert(root1, 4);
    root1 = insert(root1, 6);
    root1 = insert(root1, 8);

    root2 = insert(root2, 10);
    root2 = insert(root2, 6);
    root2 = insert(root2, 15);
    root2 = insert(root2, 3);
    root2 = insert(root2, 8);
    root2 = insert(root2, 11);
    root2 = insert(root2, 18);

    // Value of x
    int x = 16;

    // Count pairs with sum equal to x
    int count = countPairs(root1, root2, x);

    cout << "Count of pairs with sum " << x << " is "
         << count << endl;

    return 0;
}

Java

import java.util.*;

// Definition of BST node
class Node {
    int data;
    Node left, right;

    Node(int x)
    {
        data = x;
        left = right = null;
    }
}

class Main {

    // Function to insert a node in BST
    static Node insert(Node root, int x)
    {
        if (root == null)
            return new Node(x);
        if (x < root.data)
            root.left = insert(root.left, x);
        else if (x > root.data)
            root.right = insert(root.right, x);
        return root;
    }

    // Function to count pairs with sum equal to x
    static int countPairs(Node root1, Node root2, int x)
    {
        Set<Integer> set = new HashSet<>();
        Stack<Node> stack = new Stack<>();
        Node curr = root2;

        while (curr != null || !stack.empty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }

            curr = stack.pop();
            set.add(curr.data);
            curr = curr.right;
        }

        // Traverse BST 1 and search for (x-a) in the set
        int count = 0;
        curr = root1;

        while (curr != null || !stack.empty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            if (set.contains(x - curr.data))
                count++;
            curr = curr.right;
        }
        return count;
    }

    // Driver code
    public static void main(String[] args)
    {

        // Input BSTs
        Node root1 = null, root2 = null;
        root1 = insert(root1, 5);
        root1 = insert(root1, 3);
        root1 = insert(root1, 7);
        root1 = insert(root1, 2);
        root1 = insert(root1, 4);
        root1 = insert(root1, 6);
        root1 = insert(root1, 8);

        root2 = insert(root2, 10);
        root2 = insert(root2, 6);
        root2 = insert(root2, 15);
        root2 = insert(root2, 3);
        root2 = insert(root2, 8);
        root2 = insert(root2, 11);
        root2 = insert(root2, 18);

        // Value of x
        int x = 16;

        // Count pairs with sum equal to x
        int count = countPairs(root1, root2, x);
        System.out.println("Count of pairs with sum " + x
                           + " is " + count);
    }
}

Python3

# Definition of BST node
class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None

# Function to insert a node in BST


def insert(root, x):
    if root is None:
        return Node(x)
    if x < root.data:
        root.left = insert(root.left, x)
    elif x > root.data:
        root.right = insert(root.right, x)
    return root

# Function to count pairs with sum equal to x


def countPairs(root1, root2, x):
    # Set to store values of BST 2
    s = set()

    # Traverse BST 2 and insert values in the set
    st = []
    curr = root2
    while curr is not None or len(st) > 0:
        while curr is not None:
            st.append(curr)
            curr = curr.left
        curr = st.pop()
        s.add(curr.data)
        curr = curr.right

    # Traverse BST 1 and search for (x-a) in the set
    count = 0
    curr = root1
    while curr is not None or len(st) > 0:
        while curr is not None:
            st.append(curr)
            curr = curr.left
        curr = st.pop()
        if x - curr.data in s:
            count += 1
        curr = curr.right

    return count


# Driver code
if __name__ == '__main__':
    # Input BSTs
    root1, root2 = None, None
    root1 = insert(root1, 5)
    root1 = insert(root1, 3)
    root1 = insert(root1, 7)
    root1 = insert(root1, 2)
    root1 = insert(root1, 4)
    root1 = insert(root1, 6)
    root1 = insert(root1, 8)

    root2 = insert(root2, 10)
    root2 = insert(root2, 6)
    root2 = insert(root2, 15)
    root2 = insert(root2, 3)
    root2 = insert(root2, 8)
    root2 = insert(root2, 11)
    root2 = insert(root2, 18)

    # Value of x
    x = 16
    # Count pairs with sum equal to x
    print("Count of pairs with sum " + str(x) +
          " is " + str(countPairs(root1, root2, x)))

C#

using System;
using System.Collections.Generic;

// Definition of BST node
class Node {
    public int data;
    public Node left;
    public Node right;

    public Node(int x)
    {
        data = x;
        left = null;
        right = null;
    }
}

class Program {
    // Function to insert a node in BST
    static Node insert(Node root, int x)
    {
        if (root == null) {
            return new Node(x);
        }
        if (x < root.data) {
            root.left = insert(root.left, x);
        }
        else if (x > root.data) {
            root.right = insert(root.right, x);
        }
        return root;
    }

    // Function to count pairs with sum equal to x
    static int countPairs(Node root1, Node root2, int x)
    {
        // Set to store values of BST 2
        HashSet<int> s = new HashSet<int>();

        // Traverse BST 2 and insert values in the set
        Stack<Node> st = new Stack<Node>();
        Node curr = root2;
        while (curr != null || st.Count > 0) {
            while (curr != null) {
                st.Push(curr);
                curr = curr.left;
            }
            curr = st.Pop();
            s.Add(curr.data);
            curr = curr.right;
        }

        // Traverse BST 1 and search for (x-a) in the set
        int count = 0;
        curr = root1;
        st.Clear();
        while (curr != null || st.Count > 0) {
            while (curr != null) {
                st.Push(curr);
                curr = curr.left;
            }
            curr = st.Pop();
            if (s.Contains(x - curr.data)) {
                count++;
            }
            curr = curr.right;
        }

        return count;
    }

    // Driver code
    static void Main(string[] args)
    {
        // Input BSTs
        Node root1 = null, root2 = null;
        root1 = insert(root1, 5);
        root1 = insert(root1, 3);
        root1 = insert(root1, 7);
        root1 = insert(root1, 2);
        root1 = insert(root1, 4);
        root1 = insert(root1, 6);
        root1 = insert(root1, 8);

        root2 = insert(root2, 10);
        root2 = insert(root2, 6);
        root2 = insert(root2, 15);
        root2 = insert(root2, 3);
        root2 = insert(root2, 8);
        root2 = insert(root2, 11);
        root2 = insert(root2, 18);

        // Value of x
        int x = 16;

        // Count pairs with sum equal to x
        Console.WriteLine("Count of pairs with sum " + x
                          + " is "
                          + countPairs(root1, root2, x));
    }
}

Javascript

// Definition of BST node
class Node {
    constructor(x) {
        this.data = x;
        this.left = null;
        this.right = null;
    }
}

// Function to insert a node in BST
function insert(root, x) {
    if (root === null) {
        return new Node(x);
    }
    if (x < root.data) {
        root.left = insert(root.left, x);
    } else if (x > root.data) {
        root.right = insert(root.right, x);
    }
    return root;
}

// Function to count pairs with sum equal to x
function countPairs(root1, root2, x) {
    const set = new Set();
    const stack = [];
    let curr = root2;

    while (curr !== null || stack.length !== 0) {
        while (curr !== null) {
            stack.push(curr);
            curr = curr.left;
        }

        curr = stack.pop();
        set.add(curr.data);
        curr = curr.right;
    }

    // Traverse BST 1 and search for (x-a) in the set
    let count = 0;
    curr = root1;

    while (curr !== null || stack.length !== 0) {
        while (curr !== null) {
            stack.push(curr);
            curr = curr.left;
        }
        curr = stack.pop();
        if (set.has(x - curr.data)) {
            count++;
        }
        curr = curr.right;
    }
    return count;
}

// Input BSTs
let root1 = null,
    root2 = null;
root1 = insert(root1, 5);
root1 = insert(root1, 3);
root1 = insert(root1, 7);
root1 = insert(root1, 2);
root1 = insert(root1, 4);
root1 = insert(root1, 6);
root1 = insert(root1, 8);

root2 = insert(root2, 10);
root2 = insert(root2, 6);
root2 = insert(root2, 15);
root2 = insert(root2, 3);
root2 = insert(root2, 8);
root2 = insert(root2, 11);
root2 = insert(root2, 18);

// Value of x
const x = 16;

// Count pairs with sum equal to x
const count = countPairs(root1, root2, x);
console.log("Count of pairs with sum " + x + " is " + count);

Output

Count of pairs with sum 16 is 3

Time complexity: O(n1 * h2), here n1 is number of nodes in first BST and h2 is height of second BST.

Method 2: Traverse BST 1 from smallest value to node to largest. This can be achieved with the help of iterative inorder traversal. Traverse BST 2 from largest value node to smallest. This can be achieved with the help of reverse inorder traversal. Perform these two traversals simultaneously. Sum up the corresponding node’s value from both the BSTs at a particular instance of traversals. If sum == x, then increment count. If x > sum, then move to the inorder successor of the current node of BST 1, else move to the inorder predecessor of the current node of BST 2. Perform these operations until either of the two traversals gets completed.

Implementation:

C++

// C++ implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
#include <bits/stdc++.h>
using namespace std;

// structure of a node of BST
struct Node {
    int data;
    Node* left, *right;
};

// function to create and return a node of BST
Node* getNode(int data)
{
    // allocate space for the node
    Node* new_node = (Node*)malloc(sizeof(Node));

    // put in the data
    new_node->data = data;
    new_node->left = new_node->right = NULL;
}

// function to count pairs from two BSTs
// whose sum is equal to a given value x
int countPairs(Node* root1, Node* root2, int x)
{
    // if either of the tree is empty
    if (root1 == NULL || root2 == NULL)
        return 0;

    // stack 'st1' used for the inorder
    // traversal of BST 1
    // stack 'st2' used for the reverse
    // inorder traversal of BST 2
    stack<Node*> st1, st2;
    Node* top1, *top2;

    int count = 0;

    // the loop will break when either of two
    // traversals gets completed
    while (1) {

        // to find next node in inorder
        // traversal of BST 1
        while (root1 != NULL) {
            st1.push(root1);
            root1 = root1->left;
        }

        // to find next node in reverse
        // inorder traversal of BST 2
        while (root2 != NULL) {
            st2.push(root2);
            root2 = root2->right;
        }

        // if either gets empty then corresponding
        // tree traversal is completed
        if (st1.empty() || st2.empty())
            break;

        top1 = st1.top();
        top2 = st2.top();

        // if the sum of the node's is equal to 'x'
        if ((top1->data + top2->data) == x) {
            // increment count
            count++;

            // pop nodes from the respective stacks
            st1.pop();
            st2.pop();

            // insert next possible node in the
            // respective stacks
            root1 = top1->right;
            root2 = top2->left;
        }

        // move to next possible node in the
        // inorder traversal of BST 1
        else if ((top1->data + top2->data) < x) {
            st1.pop();
            root1 = top1->right;
        }

        // move to next possible node in the
        // reverse inorder traversal of BST 2
        else {
            st2.pop();
            root2 = top2->left;
        }
    }

    // required count of pairs
    return count;
}

// Driver program to test above
int main()
{
    // formation of BST 1
    Node* root1 = getNode(5); /*             5        */
    root1->left = getNode(3); /*           /   \      */
    root1->right = getNode(7); /*         3     7     */
    root1->left->left = getNode(2); /*    / \   / \    */
    root1->left->right = getNode(4); /*  2  4  6  8    */
    root1->right->left = getNode(6);
    root1->right->right = getNode(8);

    // formation of BST 2
    Node* root2 = getNode(10); /*           10         */
    root2->left = getNode(6); /*           /   \        */
    root2->right = getNode(15); /*        6     15      */
    root2->left->left = getNode(3); /*    / \   /  \     */
    root2->left->right = getNode(8); /*  3  8  11  18    */
    root2->right->left = getNode(11);
    root2->right->right = getNode(18);

    int x = 16;
    cout << "Pairs = "
         << countPairs(root1, root2, x);

    return 0;
}

Java

// Java implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
import java.util.Stack;
public class GFG {

    // structure of a node of BST
    static class Node {
        int data;
        Node left, right;

        // constructor
        public Node(int data) {
            this.data = data;
            left = null;
            right = null;
        }
    }

    static Node root1;
    static Node root2;
    // function to count pairs from two BSTs
    // whose sum is equal to a given value x
    static int countPairs(Node root1, Node root2,
                                           int x)
    {
        // if either of the tree is empty
        if (root1 == null || root2 == null)
            return 0;

        // stack 'st1' used for the inorder
        // traversal of BST 1
        // stack 'st2' used for the reverse
        // inorder traversal of BST 2
        //stack<Node*> st1, st2;
        Stack<Node> st1 = new Stack<>();
        Stack<Node> st2 = new Stack<>();
        Node top1, top2;

        int count = 0;

        // the loop will break when either of two
        // traversals gets completed
        while (true) {

            // to find next node in inorder
            // traversal of BST 1
            while (root1 != null) {
                st1.push(root1);
                root1 = root1.left;
            }

            // to find next node in reverse
            // inorder traversal of BST 2
            while (root2 != null) {
                st2.push(root2);
                root2 = root2.right;
            }

            // if either gets empty then corresponding
            // tree traversal is completed
            if (st1.empty() || st2.empty())
                break;

            top1 = st1.peek();
            top2 = st2.peek();

            // if the sum of the node's is equal to 'x'
            if ((top1.data + top2.data) == x) {
                // increment count
                count++;

                // pop nodes from the respective stacks
                st1.pop();
                st2.pop();

                // insert next possible node in the
                // respective stacks
                root1 = top1.right;
                root2 = top2.left;
            }

            // move to next possible node in the
            // inorder traversal of BST 1
            else if ((top1.data + top2.data) < x) {
                st1.pop();
                root1 = top1.right;
            }

            // move to next possible node in the
            // reverse inorder traversal of BST 2
            else {
                st2.pop();
                root2 = top2.left;
            }
        }

        // required count of pairs
        return count;
    }

    // Driver program to test above
    public static void main(String args[])
    {
        // formation of BST 1
        root1 = new Node(5);       /*             5        */
        root1.left = new Node(3); /*           /   \      */
        root1.right = new Node(7); /*         3     7     */
        root1.left.left = new Node(2); /*    / \   / \    */
        root1.left.right = new Node(4); /*  2   4 6   8    */
        root1.right.left = new Node(6);
        root1.right.right = new Node(8);

        // formation of BST 2
        root2 = new Node(10);        /*           10         */
        root2.left = new Node(6); /*           /   \        */
        root2.right = new Node(15); /*        6     15      */
        root2.left.left = new Node(3); /*    / \   /  \     */
        root2.left.right = new Node(8); /*  3  8  11  18    */
        root2.right.left = new Node(11);
        root2.right.right = new Node(18);

        int x = 16;
        System.out.println("Pairs = "
             + countPairs(root1, root2, x));
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python3 implementation to count pairs
# from two BSTs whose sum is equal to a
# given  value x

# Structure of a node of BST
class getNode:

    def __init__(self, data):

        self.data = data
        self.left = None
        self.right = None

# Function to count pairs from two BSTs
# whose sum is equal to a given value x
def countPairs(root1, root2, x):

    # If either of the tree is empty
    if (root1 == None or root2 == None):
        return 0

    # Stack 'st1' used for the inorder
    # traversal of BST 1
    # stack 'st2' used for the reverse
    # inorder traversal of BST 2
    st1 = []
    st2 = []

    count = 3

    # The loop will break when either
    # of two traversals gets completed
    while (1):

        # To find next node in inorder
        # traversal of BST 1
        while (root1 != None):
            st1.append(root1)
            root1 = root1.left

        # To find next node in reverse
        # inorder traversal of BST 2
        while (root2 != None):
            st2.append(root2)
            root2 = root2.right

        # If either gets empty then corresponding
        # tree traversal is completed
        if (len(st1) or len(st2)):
            break

        top1 = st1[len(st1) - 1]
        top2 = st2[len(st2) - 1]

        # If the sum of the node's is equal to 'x'
        if ((top1.data + top2.data) == x):

            # Increment count
            count += 1

            # Pop nodes from the respective stacks
            st1.remove(st1[len(st1) - 1])
            st2.remove(st2[len(st2) - 1])

            # Insert next possible node in the
            # respective stacks
            root1 = top1.right
            root2 = top2.left

        # Move to next possible node in the
        # inorder traversal of BST 1
        elif ((top1.data + top2.data) < x):
            st1.remove(st1[len(st1) - 1])
            root1 = top1.right

        # Move to next possible node in the
        # reverse inorder traversal of BST 2
        else:
            st2.remove(st2[len(st2) - 1])
            root2 = top2.left

    # Required count of pairs
    return count

# Driver code
if __name__ == '__main__':

    # Formation of BST 1
    '''      5
           /   \
          3     7
         / \   / \
        2   4 6   8
    '''
    root1 = getNode(5)
    root1.left = getNode(3)
    root1.right = getNode(7)
    root1.left.left = getNode(2)
    root1.left.right = getNode(4)
    root1.right.left = getNode(6)
    root1.right.right = getNode(8)

    # Formation of BST 2
    '''    10
         /   \
        6     15
       / \   /  \
      3  8  11  18
    '''
    root2 = getNode(10)
    root2.left = getNode(6)
    root2.right = getNode(15)
    root2.left.left = getNode(3)
    root2.left.right = getNode(8)
    root2.right.left = getNode(11)
    root2.right.right = getNode(18)

    x = 16

    print("Pairs = ", countPairs(root1, root2, x))

# This code is contributed by bgangwar59

C#

// C# implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
using System;
using System.Collections.Generic;

// C# implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
public class GFG
{

    // structure of a node of BST
    public class Node
    {
        public int data;
        public Node left, right;

        // constructor
        public Node(int data)
        {
            this.data = data;
            left = null;
            right = null;
        }
    }

    public static Node root1;
    public static Node root2;
    // function to count pairs from two BSTs
    // whose sum is equal to a given value x
    public static int countPairs(Node root1, Node root2, int x)
    {
        // if either of the tree is empty
        if (root1 == null || root2 == null)
        {
            return 0;
        }

        // stack 'st1' used for the inorder
        // traversal of BST 1
        // stack 'st2' used for the reverse
        // inorder traversal of BST 2
        //stack<Node*> st1, st2;
        Stack<Node> st1 = new Stack<Node>();
        Stack<Node> st2 = new Stack<Node>();
        Node top1, top2;

        int count = 0;

        // the loop will break when either of two
        // traversals gets completed
        while (true)
        {

            // to find next node in inorder
            // traversal of BST 1
            while (root1 != null)
            {
                st1.Push(root1);
                root1 = root1.left;
            }

            // to find next node in reverse
            // inorder traversal of BST 2
            while (root2 != null)
            {
                st2.Push(root2);
                root2 = root2.right;
            }

            // if either gets empty then corresponding
            // tree traversal is completed
            if (st1.Count == 0 || st2.Count == 0)
            {
                break;
            }

            top1 = st1.Peek();
            top2 = st2.Peek();

            // if the sum of the node's is equal to 'x'
            if ((top1.data + top2.data) == x)
            {
                // increment count
                count++;

                // pop nodes from the respective stacks
                st1.Pop();
                st2.Pop();

                // insert next possible node in the
                // respective stacks
                root1 = top1.right;
                root2 = top2.left;
            }

            // move to next possible node in the
            // inorder traversal of BST 1
            else if ((top1.data + top2.data) < x)
            {
                st1.Pop();
                root1 = top1.right;
            }

            // move to next possible node in the
            // reverse inorder traversal of BST 2
            else
            {
                st2.Pop();
                root2 = top2.left;
            }
        }

        // required count of pairs
        return count;
    }

    // Driver program to test above
    public static void Main(string[] args)
    {
        // formation of BST 1
        root1 = new Node(5); //             5
        root1.left = new Node(3); //           /   \
        root1.right = new Node(7); //         3     7
        root1.left.left = new Node(2); //    / \   / \
        root1.left.right = new Node(4); //  2   4 6   8
        root1.right.left = new Node(6);
        root1.right.right = new Node(8);

        // formation of BST 2
        root2 = new Node(10); //           10
        root2.left = new Node(6); //           /   \
        root2.right = new Node(15); //        6     15
        root2.left.left = new Node(3); //    / \   /  \
        root2.left.right = new Node(8); //  3  8  11  18
        root2.right.left = new Node(11);
        root2.right.right = new Node(18);

        int x = 16;
        Console.WriteLine("Pairs = " + countPairs(root1, root2, x));
    }
}

  // This code is contributed by Shrikant13

Javascript

<script>

// JavaScript implementation to count pairs
// from two BSTs whose sum is equal to a
// given  value x

// Structure of a node of BST
class getNode{

    constructor(data){
        this.data = data
        this.left = null
        this.right = null
    }

}

// Function to count pairs from two BSTs
// whose sum is equal to a given value x
function countPairs(root1, root2, x){

    // If either of the tree is empty
    if (root1 == null || root2 == null)
        return 0

    // Stack 'st1' used for the inorder
    // traversal of BST 1
    // stack 'st2' used for the reverse
    // inorder traversal of BST 2
    let st1 = []
    let st2 = []

    let count = 3

    // The loop will break when either
    // of two traversals gets completed
    while (1){

        // To find next node in inorder
        // traversal of BST 1
        while (root1 != null){
            st1.push(root1)
            root1 = root1.left
        }

        // To find next node in reverse
        // inorder traversal of BST 2
        while (root2 != null){
            st2.push(root2)
            root2 = root2.right
        }

        // If either gets empty then corresponding
        // tree traversal is completed
        if (st1.length || st2.length)
            break

        top1 = st1[st1.length - 1]
        top2 = st2[st2.length - 1]

        // If the sum of the node's is equal to 'x'
        if ((top1.data + top2.data) == x){

            // Increment count
            count += 1

            // Pop nodes from the respective stacks
            st1.pop()
            st2.pop()

            // Insert next possible node in the
            // respective stacks
            root1 = top1.right
            root2 = top2.left
        }

        // Move to next possible node in the
        // inorder traversal of BST 1
        else if ((top1.data + top2.data) < x){
            st1.pop()
            root1 = top1.right
        }

        // Move to next possible node in the
        // reverse inorder traversal of BST 2
        else{
            st2.pop()
            root2 = top2.left
        }

    }

    // Required count of pairs
    return count
}

// Driver code

// Formation of BST 1
    //      5
    //    /   \
    //   3     7
    //  / \   / \
    // 2   4 6   8

let root1 = new getNode(5)
root1.left = new getNode(3)
root1.right = new getNode(7)
root1.left.left = new getNode(2)
root1.left.right = new getNode(4)
root1.right.left = new getNode(6)
root1.right.right = new getNode(8)

// Formation of BST 2
    //       10
    //      /   \
    //     6     15
    //    / \   /  \
    //   3  8  11  18

let root2 = new getNode(10)
root2.left = new getNode(6)
root2.right = new getNode(15)
root2.left.left = new getNode(3)
root2.left.right = new getNode(8)
root2.right.left = new getNode(11)
root2.right.right = new getNode(18)

let x = 16

document.write("Pairs = ", countPairs(root1, root2, x),"</br>")

// This code is contributed by shinjanpatra

</script>

Output

Pairs = 3

Time Complexity: O(n1 + n2) 
Auxiliary Space: O(h1 + h2), Where h1 is height of first tree and h2 is height of second tree

Method 3 :

1. Recursive approach to solving this question.
2. Traverse the BST1 and for each node find the diff i.e. (x – root1.data) in BST2 and increment the count.

Implementation:

C++

// C++ implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
#include <bits/stdc++.h>
using namespace std;

// structure of a node of BST
struct Node {
    int data;
    Node *left, *right;

    // constructor
    Node(int data)
    {
        this->data = data;
        left = NULL;
        right = NULL;
    }
};

static int pairCount = 0;

void findPairs(Node* root2, int diff)
{
    if (root2 == NULL) {
        return;
    }
    if (diff > root2->data) {
        findPairs(root2->right, diff);
    }
    else {
        findPairs(root2->left, diff);
    }
    if (root2->data == diff) {
        pairCount++;
    }
}

void traverseTree(Node* root1, Node* root2, int sum)
{
    if (root1 == NULL || root2 == NULL) {
        return;
    }
    traverseTree(root1->left, root2, sum);
    traverseTree(root1->right, root2, sum);
    int diff = sum - root1->data;
    findPairs(root2, diff);
}

// function to count pairs from two BSTs
// whose sum is equal to a given value x
int countPairs(Node* root1, Node* root2, int sum)
{
    traverseTree(root1, root2, sum);
    return pairCount;
}

// Driver program to test above
int main()
{
    // formation of BST 1
    Node* root1 = new Node(5); /*             5        */
    root1->left = new Node(3); /*           /   \      */
    root1->right = new Node(7); /*         3     7     */
    root1->left->left = new Node(2); /*    / \   / \    */
    root1->left->right = new Node(4); /*  2   4 6   8 */
    root1->right->left = new Node(6);
    root1->right->right = new Node(8);

    // formation of BST 2
    Node* root2 = new Node(10); /*           10         */
    root2->left = new Node(6); /*           /   \ */
    root2->right = new Node(15); /*        6     15 */
    root2->left->left = new Node(3); /*    / \   /  \ */
    root2->left->right = new Node(8); /*  3  8  11  18    */
    root2->right->left = new Node(11);
    root2->right->right = new Node(18);

    int x = 16;
    cout << "Pairs = " << countPairs(root1, root2, x);
}

// This code is contributed by Tapesh (tapeshdua420)

Java

// Java implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
import java.util.Stack;
public class GFG {

    // structure of a node of BST
    static class Node {
        int data;
        Node left, right;

        // constructor
        public Node(int data)
        {
            this.data = data;
            left = null;
            right = null;
        }
    }

    static Node root1;
    static Node root2;

    // function to count pairs from two BSTs
    // whose sum is equal to a given value x
    public static int pairCount = 0;
    public static void traverseTree(Node root1, Node root2,
                                    int sum)
    {
        if (root1 == null || root2 == null) {
            return;
        }
        traverseTree(root1.left, root2, sum);
        traverseTree(root1.right, root2, sum);
        int diff = sum - root1.data;
        findPairs(root2, diff);
    }

    private static void findPairs(Node root2, int diff)
    {
        if (root2 == null) {
            return;
        }
        if (diff > root2.data) {
            findPairs(root2.right, diff);
        }
        else {
            findPairs(root2.left, diff);
        }
        if (root2.data == diff) {
            pairCount++;
        }
    }

    public static int countPairs(Node root1, Node root2,
                                 int sum)
    {
        traverseTree(root1, root2, sum);
        return pairCount;
    }

    // Driver program to test above
    public static void main(String args[])
    {
        // formation of BST 1
        root1 = new Node(5); /*             5        */
        root1.left = new Node(3); /*           /   \      */
        root1.right = new Node(7); /*         3     7     */
        root1.left.left = new Node(2); /*    / \   / \    */
        root1.left.right = new Node(4); /*  2   4 6   8 */
        root1.right.left = new Node(6);
        root1.right.right = new Node(8);

        // formation of BST 2
        root2 = new Node(10); /*           10         */
        root2.left = new Node(6); /*           /   \ */
        root2.right = new Node(15); /*        6     15 */
        root2.left.left = new Node(3); /*    / \   /  \ */
        root2.left.right
            = new Node(8); /*  3  8  11  18    */
        root2.right.left = new Node(11);
        root2.right.right = new Node(18);

        int x = 16;
        System.out.println("Pairs = "
                           + countPairs(root1, root2, x));
    }
}
// This code is contributed by Sujit Panda

Python3

# Python implementation to count pairs from two
# BSTs whose sum is equal to a given  value x

# structure of a node of BST
class Node:

    # constructor
    def __init__(self,data):
        self.data = data
        self.left = None
        self.right = None

root1,root2 = None,None

# def to count pairs from two BSTs
# whose sum is equal to a given value x
pairCount = 0
def traverseTree(root1,  root2, sum):

    if (root1 == None or root2 == None):
        return
    traverseTree(root1.left, root2, sum)
    traverseTree(root1.right, root2, sum)
    diff = sum - root1.data
    findPairs(root2, diff)

def findPairs(root2 , diff):

    global pairCount

    if (root2 == None):
        return

    if (diff > root2.data) :
        findPairs(root2.right, diff)
    else :
        findPairs(root2.left, diff)
    if (root2.data == diff):
        pairCount += 1

def countPairs(root1, root2, sum):
    global pairCount

    traverseTree(root1, root2, sum)
    return pairCount

# Driver program to test above

# formation of BST 1
root1 = Node(5)
root1.left = Node(3)
root1.right = Node(7)
root1.left.left = Node(2)
root1.left.right = Node(4)
root1.right.left = Node(6)
root1.right.right = Node(8)

# formation of BST 2
root2 = Node(10)
root2.left = Node(6)
root2.right = Node(15)
root2.left.left = Node(3)
root2.left.right = Node(8)
root2.right.left = Node(11)
root2.right.right = Node(18)

x = 16
print(f"Pairs = {countPairs(root1, root2, x)}")

# This code is contributed by shinjanpatra

C#

// C# implementation to count pairs from two
// BSTs whose sum is equal to a given  value x
using System;

using System.Collections.Generic;
public class GFG {

    // structure of a node of BST
    public class Node {
        public int data;
        public Node left, right;

        // constructor
        public Node(int data)
        {
            this.data = data;
            left = null;
            right = null;
        }
    }

    static Node root1;
    static Node root2;

    // function to count pairs from two BSTs
    // whose sum is equal to a given value x
    public static int pairCount = 0;
    public static void traverseTree(Node root1, Node root2,
                                    int sum)
    {
        if (root1 == null || root2 == null) {
            return;
        }
        traverseTree(root1.left, root2, sum);
        traverseTree(root1.right, root2, sum);
        int diff = sum - root1.data;
        findPairs(root2, diff);
    }

    private static void findPairs(Node root2, int diff)
    {
        if (root2 == null) {
            return;
        }
        if (diff > root2.data) {
            findPairs(root2.right, diff);
        }
        else {
            findPairs(root2.left, diff);
        }
        if (root2.data == diff) {
            pairCount++;
        }
    }

    public static int countPairs(Node root1, Node root2,
                                 int sum)
    {
        traverseTree(root1, root2, sum);
        return pairCount;
    }

    // Driver program to test above
    public static void Main(String []args)
    {
        // formation of BST 1
        root1 = new Node(5); /*             5        */
        root1.left = new Node(3); /*           /   \      */
        root1.right = new Node(7); /*         3     7     */
        root1.left.left = new Node(2); /*    / \   / \    */
        root1.left.right = new Node(4); /*  2   4 6   8 */
        root1.right.left = new Node(6);
        root1.right.right = new Node(8);

        // formation of BST 2
        root2 = new Node(10); /*           10         */
        root2.left = new Node(6); /*           /   \ */
        root2.right = new Node(15); /*        6     15 */
        root2.left.left = new Node(3); /*    / \   /  \ */
        root2.left.right
            = new Node(8); /*  3  8  11  18    */
        root2.right.left = new Node(11);
        root2.right.right = new Node(18);

        int x = 16;
        Console.WriteLine("Pairs = "
                           + countPairs(root1, root2, x));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// javascript implementation to count pairs from two
// BSTs whose sum is equal to a given  value x

    // structure of a node of BST
     class Node {

        // constructor
        constructor(data)
        {
            this.data = data;
            this.left = null;
            this.right = null;
        }
    }

    var root1;
    var root2;

    // function to count pairs from two BSTs
    // whose sum is equal to a given value x
    var pairCount = 0;
    function traverseTree(root1,  root2,
                                     sum)
    {
        if (root1 == null || root2 == null) {
            return;
        }
        traverseTree(root1.left, root2, sum);
        traverseTree(root1.right, root2, sum);
        var diff = sum - root1.data;
        findPairs(root2, diff);
    }

     function findPairs(root2 , diff)
    {
        if (root2 == null) {
            return;
        }
        if (diff > root2.data) {
            findPairs(root2.right, diff);
        }
        else {
            findPairs(root2.left, diff);
        }
        if (root2.data == diff) {
            pairCount++;
        }
    }

    function countPairs(root1,  root2,
                                  sum)
    {
        traverseTree(root1, root2, sum);
        return pairCount;
    }

    // Driver program to test above

        // formation of BST 1
        root1 = new Node(5); /*             5        */
        root1.left = new Node(3); /*           /   \      */
        root1.right = new Node(7); /*         3     7     */
        root1.left.left = new Node(2); /*    / \   / \    */
        root1.left.right = new Node(4); /*  2   4 6   8 */
        root1.right.left = new Node(6);
        root1.right.right = new Node(8);

        // formation of BST 2
        root2 = new Node(10); /*           10         */
        root2.left = new Node(6); /*           /   \ */
        root2.right = new Node(15); /*        6     15 */
        root2.left.left = new Node(3); /*    / \   /  \ */
        root2.left.right
            = new Node(8); /*  3  8  11  18    */
        root2.right.left = new Node(11);
        root2.right.right = new Node(18);

        var x = 16;
        document.write("Pairs = "
                           + countPairs(root1, root2, x));

// This code is contributed by Rajput-Ji
</script>

Output

Pairs = 3

Time Complexity: O(n1 * n2), As for every node in the BST1 we need to traverse the BST2 and check if it is equal to diff or not.
Auxiliary Space: O(h1 + h2), Here h1 is the height of BST1 and h2 is the height of the BST2.

Method 4 : Using BinarySearch Tree Iterator ( A more general way of doing this )

Create two class one as BSTIterator1 and other as BSTIterator2. These two class corresponds to inOrder and reverse inOrder traversal respectively.

Each class will have three methods:

• hasNext : will return true when traversal is not yet completed
• next : will move the pointer to the next node
• peek : will return current node in the traversal

After creating two such classes, simple run the iterator while both have next node and find the sum. If sum == x, increment the next pointer of iterator1 and iterator2 and if sum > x ,increment the next pointer of iterator2 else increment the next pointer of iterator1 i.e when sum < x.

Implementation:

C++

// C++ Program to Count pairs from two BSTs whose sum is
// equal to a given value x using BinarySearch Tree Iterator
#include <bits/stdc++.h>
using namespace std;

class Node {
public:
    int data;
    Node* left;
    Node* right;
    Node(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};

// inorder successor iterator
class BSTIterator1 {
public:
    stack<Node*> s1;
    Node* root1;
    bool hasPeeked = false;
    BSTIterator1(Node* root) { this->root1 = root; }
    bool hasNext()
    {
        if (!s1.empty() || root1 != NULL)
            return true;
        return false;
    }
    Node* peek()
    {
        if (!hasNext())
            return NULL;
        while (root1 != NULL) {
            s1.push(root1);
            root1 = root1->left;
            hasPeeked = true;
        }
        return s1.top();
    }
    int next()
    {
        if (!hasNext())
            return -1;
        if (!hasPeeked)
            peek();
        hasPeeked = false;
        root1 = s1.top();
        s1.pop();
        Node* temp = root1;
        root1 = root1->right;
        return temp->data;
    }
};

// inorder predecessor iterator
class BSTIterator2 {
public:
    stack<Node*> s1;
    Node* root1;
    bool hasPeeked = false;
    BSTIterator2(Node* root) { this->root1 = root; }
    bool hasNext()
    {
        if (!s1.empty() || root1 != NULL)
            return true;
        return false;
    }
    Node* peek()
    {
        if (!hasNext())
            return NULL;
        while (root1 != NULL) {
            s1.push(root1);
            root1 = root1->right;
            hasPeeked = true;
        }
        return s1.top();
    }
    int next()
    {
        if (!hasNext())
            return -1;
        if (!hasPeeked)
            peek();
        hasPeeked = false;
        root1 = s1.top();
        s1.pop();
        Node* temp = root1;
        root1 = root1->left;
        return temp->data;
    }
};

int countPairs(Node* r1, Node* r2, int x)
{

    BSTIterator1* it1 = new BSTIterator1(r1);
    BSTIterator2* it2 = new BSTIterator2(r2);
    int count = 0;
    while (it1->hasNext() && it2->hasNext()) {
        Node* n1 = it1->peek();
        Node* n2 = it2->peek();
        int sum = n1->data + n2->data;
        if (sum == x) {
            count++;
            it1->next();
            it2->next();
        }
        else if (sum > x) {
            it2->next();
        }
        else {
            it1->next();
        }
    }
    return count;
}

// Driver program to test above
int main()
{
    Node *root1, *root2;
    // formation of BST 1
    root1 = new Node(5); /*                     5        */
    root1->left = new Node(3); /*           /   \      */
    root1->right = new Node(7); /*         3     7     */
    root1->left->left = new Node(2); /*    / \   / \    */
    root1->left->right = new Node(4); /*  2   4 6   8   */
    root1->right->left = new Node(6);
    root1->right->right = new Node(8);

    // formation of BST 2
    root2 = new Node(10); /*                   10      */
    root2->left = new Node(6); /*           /   \    */
    root2->right = new Node(15); /*        6     15  */
    root2->left->left = new Node(3); /*    / \   /  \ */
    root2->left->right = new Node(8); /*  3  8  11  18    */
    root2->right->left = new Node(11);
    root2->right->right = new Node(18);

    int x = 16;
    cout << "Pairs = " << countPairs(root1, root2, x)
         << endl;
}

// This code is contributed by Tapesh (tapeshdua420)

Java

import java.util.*;
class Node{
    int data;
      Node left, right;
       public Node(int data){
        this.data = data;
          this.left = null;
          this.right = null;
    }
}
// inorder successor iterator
class BSTIterator1{
    Stack<Node> s1 = new Stack<>();
    Node root1;
    boolean hasPeeked = false;
    public BSTIterator1(Node root){
        this.root1 = root;
    }
    public boolean hasNext(){
        if(!s1.isEmpty() || root1!=null)
            return true;
        return false;
    }
    public Node peek(){
        if(!hasNext())
            return null;
        while(root1!=null){
            s1.push(root1);
            root1 = root1.left;
            hasPeeked = true;
        }
        return s1.peek();
    }
    public int next(){
        if(!hasNext())
            return -1;
        if(!hasPeeked)
            peek();
        hasPeeked = false;
        root1 = s1.pop();
        Node temp = root1;
        root1 = root1.right;
        return temp.data;
    }
}
// inorder predecessor iterator
class BSTIterator2{
    Stack<Node> s1 = new Stack<>();
    Node root1;
    boolean hasPeeked = false;
    public BSTIterator2(Node root){
        this.root1 = root;
    }
    public boolean hasNext(){
        if(!s1.isEmpty() || root1!=null)
            return true;
        return false;
    }
    public Node peek(){
        if(!hasNext())
            return null;
        while(root1!=null){
            s1.push(root1);
            root1 = root1.right;
            hasPeeked = true;
        }
        return s1.peek();
    }
    public int next(){
        if(!hasNext())
            return -1;
        if(!hasPeeked)
            peek();
        hasPeeked = false;
        root1 = s1.pop();
        Node temp = root1;
        root1 = root1.left;
        return temp.data;
    }
}
class GfG
{
    public static int countPairs(Node r1, Node r2, int x)
    {

        BSTIterator1 it1 = new BSTIterator1(r1);
        BSTIterator2 it2 = new BSTIterator2(r2);
        int count = 0;
        while(it1.hasNext() && it2.hasNext()){
            Node n1 = it1.peek();
            Node n2 = it2.peek();
            int sum = n1.data+n2.data;
            if(sum == x){
                count++;
                it1.next();
                it2.next();
            }
            else if(sum > x){
                it2.next();
            }else{
                it1.next();
            }
        }
        return count;

    }
   // Driver program to test above
    public static void main(String args[])
    {
        Node root1, root2;
          // formation of BST 1
        root1 = new Node(5); /*                     5        */
        root1.left = new Node(3); /*           /   \      */
        root1.right = new Node(7); /*         3     7     */
        root1.left.left = new Node(2); /*    / \   / \    */
        root1.left.right = new Node(4); /*  2   4 6   8   */
        root1.right.left = new Node(6);
        root1.right.right = new Node(8);

        // formation of BST 2
        root2 = new Node(10); /*                   10      */
        root2.left = new Node(6); /*           /   \    */
        root2.right = new Node(15); /*        6     15  */
        root2.left.left = new Node(3); /*    / \   /  \ */
        root2.left.right
            = new Node(8); /*  3  8  11  18    */
        root2.right.left = new Node(11);
        root2.right.right = new Node(18);

        int x = 16;
        System.out.println("Pairs = "
                           + countPairs(root1, root2, x));
    }
}

Python3

class Node:
    data = 0
    left = None
    right = None

    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# inorder successor iterator
class BSTIterator1:
    s1 = []
    root1 = None
    hasPeeked = False

    def __init__(self, root):
        self.root1 = root

    def hasNext(self):
        if (not (len(self.s1) == 0) or self.root1 != None):
            return True
        return False

    def peek(self):
        if (not self.hasNext()):
            return None
        while (self.root1 != None):
            self.s1.append(self.root1)
            self.root1 = self.root1.left
            self.hasPeeked = True
        return self.s1[-1]

    def next(self):
        if (not self.hasNext()):
            return -1
        if (not self.hasPeeked):
            self.peek()
        self.hasPeeked = False
        self.root1 = self.s1.pop()
        temp = self.root1
        self.root1 = self.root1.right
        return temp.data

# inorder predecessor iterator
class BSTIterator2:
    s1 = []
    root1 = None
    hasPeeked = False

    def __init__(self, root):
        self.root1 = root

    def hasNext(self):
        if (not (len(self.s1) == 0) or self.root1 != None):
            return True
        return False

    def peek(self):
        if (not self.hasNext()):
            return None
        while (self.root1 != None):
            self.s1.append(self.root1)
            self.root1 = self.root1.right
            self.hasPeeked = True
        return self.s1[-1]

    def next(self):
        if (not self.hasNext()):
            return -1
        if (not self.hasPeeked):
            self.peek()
        self.hasPeeked = False
        self.root1 = self.s1.pop()
        temp = self.root1
        self.root1 = self.root1.left
        return temp.data

class GfG:
    @staticmethod
    def countPairs(r1,  r2,  x):
        it1 = BSTIterator1(r1)
        it2 = BSTIterator2(r2)
        count = 0
        while (it1.hasNext() and it2.hasNext()):
            n1 = it1.peek()
            n2 = it2.peek()
            sum = n1.data + n2.data
            if (sum == x):
                count += 1
                it1.next()
                it2.next()
            elif(sum > x):
                it2.next()
            else:
                it1.next()
        return count

    # Driver program to test above
    @staticmethod
    def main(args):
        root1 = None
        root2 = None

        # formation of BST 1
        root1 = Node(5)
        #                     5
        root1.left = Node(3)
        #           /   \
        root1.right = Node(7)
        #         3     7
        root1.left.left = Node(2)
        #    / \   / \
        root1.left.right = Node(4)
        #  2   4 6   8
        root1.right.left = Node(6)
        root1.right.right = Node(8)
        # formation of BST 2
        root2 = Node(10)
        #                   10
        root2.left = Node(6)
        #           /   \
        root2.right = Node(15)
        #        6     15
        root2.left.left = Node(3)
        #    / \   /  \
        root2.left.right = Node(8)
        #  3  8  11  18
        root2.right.left = Node(11)
        root2.right.right = Node(18)
        x = 16
        print("Pairs = " + str(GfG.countPairs(root1, root2, x)))

if __name__ == "__main__":
    GfG.main([])

# This code is contributed by mukulsomukesh

C#

using System;
using System.Collections.Generic;

public class Node{
  public int data;
  public Node left, right;
  public Node(int data){
    this.data = data;
    this.left = null;
    this.right = null;
  }
}

// inorder successor iterator
public class BSTIterator1{
  public Stack<Node> s1 = new Stack<Node>();
  public Node root1;
  public bool hasPeeked = false;
  public BSTIterator1(Node root){
    this.root1 = root;
  }
  public bool hasNext(){
    if(s1.Count != 0 || root1 != null)
      return true;
    return false;
  }
  public Node peek(){
    if(!hasNext())
      return null;
    while(root1 != null){
      s1.Push(root1);
      root1 = root1.left;
      hasPeeked = true;
    }
    return s1.Peek();
  }
  public int next(){
    if(!hasNext())
      return -1;
    if(!hasPeeked)
      peek();
    hasPeeked = false;
    root1 = s1.Pop();
    Node temp = root1;
    root1 = root1.right;
    return temp.data;
  }
}
// inorder predecessor iterator
public class BSTIterator2{
  public Stack<Node> s1 = new Stack<Node>();
  public Node root1;
  public bool hasPeeked = false;
  public BSTIterator2(Node root){
    this.root1 = root;
  }
  public bool hasNext(){
    if(s1.Count != 0 || root1 != null)
      return true;
    return false;
  }
  public Node peek(){
    if(!hasNext())
      return null;
    while(root1 != null){
      s1.Push(root1);
      root1 = root1.right;
      hasPeeked = true;
    }
    return s1.Peek();
  }
  public int next(){
    if(!hasNext())
      return -1;
    if(!hasPeeked)
      peek();
    hasPeeked = false;
    root1 = s1.Pop();
    Node temp = root1;
    root1 = root1.left;
    return temp.data;
  }
}
public class GfG
{
  public static int countPairs(Node r1, Node r2, int x)
  {

    BSTIterator1 it1 = new BSTIterator1(r1);
    BSTIterator2 it2 = new BSTIterator2(r2);
    int count = 0;
    while(it1.hasNext() && it2.hasNext()){
      Node n1 = it1.peek();
      Node n2 = it2.peek();
      int sum = n1.data+n2.data;
      if(sum == x){
        count++;
        it1.next();
        it2.next();
      }
      else if(sum > x){
        it2.next();
      }else{
        it1.next();
      }
    }
    return count;

  }

  // Driver program to test above
  public static void Main(String []args)
  {
    Node root1, root2;

    // formation of BST 1
    root1 = new Node(5); /*                     5        */
    root1.left = new Node(3); /*           /   \      */
    root1.right = new Node(7); /*         3     7     */
    root1.left.left = new Node(2); /*    / \   / \    */
    root1.left.right = new Node(4); /*  2   4 6   8   */
    root1.right.left = new Node(6);
    root1.right.right = new Node(8);

    // formation of BST 2
    root2 = new Node(10); /*                   10      */
    root2.left = new Node(6); /*           /   \    */
    root2.right = new Node(15); /*        6     15  */
    root2.left.left = new Node(3); /*    / \   /  \ */
    root2.left.right
      = new Node(8); /*  3  8  11  18    */
    root2.right.left = new Node(11);
    root2.right.right = new Node(18);

    int x = 16;
    Console.WriteLine("Pairs = "
                      + countPairs(root1, root2, x));
  }
}

// This code is contributed by Rajput-Ji

Javascript

class Node {
    constructor(data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

class BSTIterator1 {
    constructor(root) {
        this.s1 = [];
        this.root1 = root;
        this.hasPeeked = false;
    }

    hasNext() {
        if (this.s1.length === 0 && this.root1 === null) {
            return false;
        }
        return true;
    }

    peek() {
        if (!this.hasNext()) {
            return null;
        }
        while (this.root1 !== null) {
            this.s1.push(this.root1);
            this.root1 = this.root1.left;
            this.hasPeeked = true;
        }
        return this.s1[this.s1.length - 1];
    }

    next() {
        if (!this.hasNext()) {
            return -1;
        }
        if (!this.hasPeeked) {
            this.peek();
        }
        this.hasPeeked = false;
        let temp = this.s1.pop();
        this.root1 = temp.right;
        return temp.data;
    }
}

class BSTIterator2 {
    constructor(root) {
        this.s1 = [];
        this.root1 = root;
        this.hasPeeked = false;
    }

    hasNext() {
        if (this.s1.length === 0 && this.root1 === null) {
            return false;
        }
        return true;
    }

    peek() {
        if (!this.hasNext()) {
            return null;
        }
        while (this.root1 !== null) {
            this.s1.push(this.root1);
            this.root1 = this.root1.right;
            this.hasPeeked = true;
        }
        return this.s1[this.s1.length - 1];
    }

    next() {
        if (!this.hasNext()) {
            return -1;
        }
        if (!this.hasPeeked) {
            this.peek();
        }
        this.hasPeeked = false;
        let temp = this.s1.pop();
        this.root1 = temp.left;
        return temp.data;
    }
}

class GfG {
    static countPairs(r1, r2, x) {
        let it1 = new BSTIterator1(r1);
        let it2 = new BSTIterator2(r2);
        let count = 0;
        while (it1.hasNext() && it2.hasNext()) {
        let n1 = it1.peek();
        let n2 = it2.peek();
        let sum = n1.data + n2.data;
        if (sum == x) {
           count++;
           it1.next();
           it2.next();
        } else if (sum > x) {
            it2.next();
            } else {
           it1.next();
        }
       }
       return count;
    }
        }

// Driver program to test above function
let root1 = null;
let root2 = null;

// formation of BST 1
root1 = new Node(5);
// 5
root1.left = new Node(3);
// /
root1.right = new Node(7);
// 3 7
root1.left.left = new Node(2);
root1.left.right = new Node(4);
// 2 4 6 8
root1.right.left = new Node(6);
root1.right.right = new Node(8);

// formation of BST 2
root2 = new Node(10);
// 10
root2.left = new Node(6);
// /
root2.right = new Node(15);
// 6 15
root2.left.left = new Node(3);
root2.left.right = new Node(8);
console.log(GfG.countPairs(root1, root2, 10));

Output

Pairs = 3

Time Complexity: O(n1 + n2)
Auxiliary Space: O(h1 + h2), Where h1 is height of first tree and h2 is height of second tree