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# Find a pair with given sum in BST

Given a BST and a sum, find if there is a pair with the given sum.

Example:

Input: sum = 28, given BST

Output: Pair is found (16, 12)

## Pair with given sum using Hashing

The idea is based on Hashing. We traverse binary search tree by inorder way and insert node’s value into a set. Also check for any node, difference between given sum and node’s value in set, if it is found then pair exists otherwise it doesn’t exist.

Follow the steps mentioned below to implement the idea:

• Traverse the tree, while traversing store the value of a node in the set
• If for a current node with value x, there exists a y for which x + y = sum then check it using set and return the pair.

Below is the implementation of the above approach:

## C++

 // CPP program to find a pair with // given sum using hashing #include using namespace std;   struct Node {     int data;     struct Node *left, *right; };   Node* NewNode(int data) {     Node* temp = (Node*)malloc(sizeof(Node));     temp->data = data;     temp->left = NULL;     temp->right = NULL;     return temp; }   Node* insert(Node* root, int key) {     if (root == NULL)         return NewNode(key);     if (key < root->data)         root->left = insert(root->left, key);     else         root->right = insert(root->right, key);     return root; }   bool findpairUtil(Node* root, int sum,                   unordered_set& set) {     if (root == NULL)         return false;       if (findpairUtil(root->left, sum, set))         return true;       if (set.find(sum - root->data) != set.end()) {         cout << "Pair is found (" << sum - root->data              << ", " << root->data << ")" << endl;         return true;     }     else         set.insert(root->data);       return findpairUtil(root->right, sum, set); }   void findPair(Node* root, int sum) {     unordered_set set;     if (!findpairUtil(root, sum, set))         cout << "Pairs do not exit" << endl; }   // Driver code int main() {     Node* root = NULL;     root = insert(root, 15);     root = insert(root, 10);     root = insert(root, 20);     root = insert(root, 8);     root = insert(root, 12);     root = insert(root, 16);     root = insert(root, 25);     root = insert(root, 10);       int sum = 28;     findPair(root, sum);       return 0; }

## Java

 // THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002) // Java code to find a pair with given sum // using hashing approach import java.util.*;   public class GFG {     // node class structure     static class Node {         int data;         Node left, right;     };         // utility function that returns the new node     static Node NewNode(int data){         Node temp = new Node();         temp.data = data;         temp.left = null;         temp.right = null;         return temp;     }         // inserting node at correct position in BST     static Node insert(Node root, int key){         if (root == null) return NewNode(key);         if (key < root.data) root.left = insert(root.left, key);         else root.right = insert(root.right, key);         return root;     }       static boolean findpairUtil(Node root, int sum, HashSet set){           // base case         if (root == null) return false;           if (findpairUtil(root.left, sum, set)) return true;           if (set.contains(sum - root.data)){             System.out.println("Pair is found (" + (sum - root.data) + ", " + root.data + ")");             return true;         }         else set.add(root.data);           return findpairUtil(root.right, sum, set);     }       static void findPair(Node root, int sum){         HashSet set = new HashSet();         if (!findpairUtil(root, sum, set))             System.out.print("Pairs do not exit \n");     }       // Driver code to test above function     public static void main(String[] args){         Node root = null;         root = insert(root, 15);         root = insert(root, 10);         root = insert(root, 20);         root = insert(root, 8);         root = insert(root, 12);         root = insert(root, 16);         root = insert(root, 25);         root = insert(root, 10);           int sum = 28;         findPair(root, sum);     } }

## Python3

 # Python3 program to find a pair with # given sum using hashing import sys import math     class Node:     def __init__(self, data):         self.data = data         self.left = None         self.right = None     def insert(root, data):     if root is None:         return Node(data)     if(data < root.data):         root.left = insert(root.left, data)     if(data > root.data):         root.right = insert(root.right, data)     return root     def findPairUtil(root, summ, unsorted_set):     if root is None:         return False     if findPairUtil(root.left, summ, unsorted_set):         return True     if unsorted_set and (summ-root.data) in unsorted_set:         print("Pair is Found ({},{})".format(summ-root.data, root.data))         return True     else:         unsorted_set.add(root.data)       return findPairUtil(root.right, summ, unsorted_set)     def findPair(root, summ):     unsorted_set = set()     if(not findPairUtil(root, summ, unsorted_set)):         print("Pair do not exist!")     # Driver code if __name__ == '__main__':     root = None     root = insert(root, 15)     root = insert(root, 10)     root = insert(root, 20)     root = insert(root, 8)     root = insert(root, 12)     root = insert(root, 16)     root = insert(root, 25)     root = insert(root, 10)     summ = 28     findPair(root, summ)   # This code is contributed by Vikash Kumar 37

## C#

 // C# program to find a pair with // given sum using hashing using System; using System.Collections.Generic;   class GFG {       class Node {         public int data;         public Node left, right;     };       static Node NewNode(int data)     {         Node temp = new Node();         temp.data = data;         temp.left = null;         temp.right = null;         return temp;     }       static Node insert(Node root, int key)     {         if (root == null)             return NewNode(key);         if (key < root.data)             root.left = insert(root.left, key);         else             root.right = insert(root.right, key);         return root;     }       static bool findpairUtil(Node root, int sum,                              HashSet set)     {         if (root == null)             return false;           if (findpairUtil(root.left, sum, set))             return true;           if (set.Contains(sum - root.data)) {             Console.WriteLine("Pair is found ("                               + (sum - root.data) + ", "                               + root.data + ")");             return true;         }         else             set.Add(root.data);           return findpairUtil(root.right, sum, set);     }       static void findPair(Node root, int sum)     {         HashSet set = new HashSet();         if (!findpairUtil(root, sum, set))             Console.Write("Pairs do not exit"                           + "\n");     }       // Driver code     public static void Main(String[] args)     {         Node root = null;         root = insert(root, 15);         root = insert(root, 10);         root = insert(root, 20);         root = insert(root, 8);         root = insert(root, 12);         root = insert(root, 16);         root = insert(root, 25);         root = insert(root, 10);           int sum = 28;         findPair(root, sum);     } }   // This code is contributed by Rajput-Ji

## Javascript

 // JavaScript program to find a pair with // given sum using hashing class Node {     constructor()     {         this.data = 0;         this.left = null;         this.right = null;     } }; function NewNode(data) {     var temp = new Node();     temp.data = data;     temp.left = null;     temp.right = null;     return temp; } function insert(root, key) {     if (root == null)         return NewNode(key);     if (key < root.data)         root.left = insert(root.left, key);     else         root.right = insert(root.right, key);     return root; } function findpairUtil(root, sum, set) {     if (root == null)         return false;     if (findpairUtil(root.left, sum, set))         return true;     if (set.has(sum - root.data)) {         console.log("Pair is found ("                           + (sum - root.data) + ", "                           + root.data + ")
");         return true;     }     else         set.add(root.data);     return findpairUtil(root.right, sum, set); } function findPair(root, sum) {     var set = new Set();     if (!findpairUtil(root, sum, set))         console.log("Pairs do not exit"                       + "\n"); } // Driver code var root = null; root = insert(root, 15); root = insert(root, 10); root = insert(root, 20); root = insert(root, 8); root = insert(root, 12); root = insert(root, 16); root = insert(root, 25); root = insert(root, 10); var sum = 28; findPair(root, sum);

Output

Pair is found (12, 16)

Time Complexity: O(N)
Auxiliary Space: O(N)

## Pair with a given sum using Two pointers

Key Idea:

The main idea is to find the Inorder Traversal of the BST and store it in a vector. We know that Inorder traversal of BST will be in sorted order.Now we will Apply Two pointers Technique to find that whether there exist two elements in the vector that sums up to the given value.

Algorithm:

1. First find the Inorder traversal of the Given BST and store it in a vector  (Let v).

2. Take two pointers i and j. Keep i at the start of v and j at the end of the v.

Now, if sum of elements at the ith index and jth index is greater that the given element then decrement j,

if sum of elements at the ith index and jth index is less that the given element then increment i,

else, these two elements are our required answer.

Below is the implementation of the above approach:

## C++

 #include using namespace std;   struct TreeNode {     int val;     struct TreeNode *left, *right; };   TreeNode* NewNode(int data) {     TreeNode* temp = (TreeNode*)malloc(sizeof(TreeNode));     temp->val = data;     temp->left = NULL;     temp->right = NULL;     return temp; }   TreeNode* insert(TreeNode* root, int key) {     if (root == NULL)         return NewNode(key);     if (key < root->val)         root->left = insert(root->left, key);     else         root->right = insert(root->right, key);     return root; }   void inorder(TreeNode* root, vector& v) {     if (root == NULL) {         return;     }     inorder(root->left, v);     v.push_back(root->val);     inorder(root->right, v); }   pair findTarget(TreeNode* root, int k) {       vector v;     inorder(root, v);     int n = v.size();     int i = 0;     int j = n - 1;     while (j > i) {         if (v[i] + v[j] == k) {             return { v[i], v[j] };         }         else if (v[i] + v[j] > k) {             j--;         }         else {             i++;         }     }     return { -1, -1 }; }   int main() {       TreeNode* root = NULL;     root = insert(root, 15);     root = insert(root, 10);     root = insert(root, 20);     root = insert(root, 8);     root = insert(root, 12);     root = insert(root, 16);     root = insert(root, 25);     root = insert(root, 10);       int k = 28;       auto a = findTarget(root, k);     cout << a.first << " " << a.second << endl; }

## Java

 import java.util.*;   public class GFG {     static class pair {     int first, second;     pair(int f, int s)     {       first = f;       second = s;     }   }   static class NewNode {     int val;     NewNode left, right;       NewNode(int data)     {       val = data;       left = null;       right = null;     }   };     static NewNode insert(NewNode root, int key)   {     if (root == null)       return new NewNode(key);     if (key < root.val)       root.left = insert(root.left, key);     else       root.right = insert(root.right, key);     return root;   }     static void inorder(NewNode root, ArrayList v)   {     if (root == null) {       return;     }     inorder(root.left, v);     v.add(root.val);     inorder(root.right, v);   }     static pair findTarget(NewNode root, int k)   {       ArrayList v = new ArrayList<>();     inorder(root, v);     int n = v.size();     int i = 0;     int j = n - 1;     while (j > i) {       if (v.get(i) + v.get(j) == k) {         return new pair(v.get(i), v.get(j));       }       else if (v.get(i) + v.get(j) > k) {         j--;       }       else {         i++;       }     }     return new pair(-1, -1);   }     public static void main(String[] args)   {     NewNode root = null;     root = insert(root, 15);     root = insert(root, 10);     root = insert(root, 20);     root = insert(root, 8);     root = insert(root, 12);     root = insert(root, 16);     root = insert(root, 25);     root = insert(root, 10);       int k = 28;       pair a = findTarget(root, k);     System.out.println(a.first + " " + a.second);   } }   // This code is contributed by Karandeep1234

## Python3

 class TreeNode:     def __init__(self, val):         self.val = val         self.left = None         self.right = None   def NewNode(data):     temp = TreeNode(data)     return temp   def insert(root, key):     if root is None:         return NewNode(key)     if key < root.val:         root.left = insert(root.left, key)     else:         root.right = insert(root.right, key)     return root   def inorder(root, v):     if root is None:         return     inorder(root.left, v)     v.append(root.val)     inorder(root.right, v)   def findTarget(root, k):     v = []     inorder(root, v)     n = len(v)     i = 0     j = n - 1     while j > i:         if v[i] + v[j] == k:             return (v[i], v[j     ])         elif v[i] + v[j] > k:             j -= 1         else:             i += 1     return (-1, -1)   if __name__ == '__main__':     root = None     root = insert(root, 15)     root = insert(root, 10)     root = insert(root, 20)     root = insert(root, 8)     root = insert(root, 12)     root = insert(root, 16)     root = insert(root, 25)     root = insert(root, 10)       k = 28       a = findTarget(root, k)     print(a[0], a[1])

## C#

 using System; using System.Collections.Generic;   public class GFG {   public class pair   {     public int first, second;     public pair(int f, int s)     {       first = f;       second = s;     }   }   public class NewNode   {     public int val;     public NewNode left, right;     public NewNode(int data)     {       val = data;       left = null;       right = null;     }   };     public static NewNode insert(NewNode root, int key)   {     if (root == null)       return new NewNode(key);     if (key < root.val)       root.left = insert(root.left, key);     else       root.right = insert(root.right, key);     return root;   }     public static void inorder(NewNode root, List v)   {     if (root == null)     {       return;     }     inorder(root.left, v);     v.Add(root.val);     inorder(root.right, v);   }     public static pair findTarget(NewNode root, int k)   {       List v = new List();     inorder(root, v);     int n = v.Count;     int i = 0;     int j = n - 1;     while (j > i)     {       if (v[i] + v[j] == k)       {         return new pair(v[i], v[j]);       }       else if (v[i] + v[j] > k)       {         j--;       }       else       {         i++;       }     }     return new pair(-1, -1);   }     public static void Main (String[] args)   {     NewNode root = null;     root = insert(root, 15);     root = insert(root, 10);     root = insert(root, 20);     root = insert(root, 8);     root = insert(root, 12);     root = insert(root, 16);     root = insert(root, 25);     root = insert(root, 10);       int k = 28;       pair a = findTarget(root, k);     Console.WriteLine(a.first + " " + a.second);   } }   // This code is contributed by Ajax

## Javascript

 class TreeNode {     constructor(val) {         this.val = val;         this.left = null;         this.right = null;     } }   function NewNode(data) {     let temp = new TreeNode(data);     return temp; }   function insert(root, key) {     if (root === null) {         return NewNode(key);     }     if (key < root.val) {         root.left = insert(root.left, key);     } else {         root.right = insert(root.right, key);     }     return root; }   function inorder(root, v) {     if (root === null) {         return;     }     inorder(root.left, v);     v.push(root.val);     inorder(root.right, v); }   function findTarget(root, k) {     let v = [];     inorder(root, v);     let n = v.length;     let i = 0;     let j = n - 1;     while (j > i) {         if (v[i] + v[j] === k) {             return [v[i], v[j]];         } else if (v[i] + v[j] > k) {             j -= 1;         } else {             i += 1;         }     }     return [-1, -1]; }   let root = null; root = insert(root, 15); root = insert(root, 10); root = insert(root, 20); root = insert(root, 8); root = insert(root, 12); root = insert(root, 16); root = insert(root, 25); root = insert(root, 10);   let k = 28;   let a = findTarget(root, k); console.log(a[0], a[1]);

Output

8 20

Time Complexity: O(n) (n = number of nodes)

Auxiliary Space: O(n)

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