Find if there is a triplet in a Balanced BST that adds to zero

Given a Balanced Binary Search Tree (BST), the task is to write a function isTripletPresent() which returns true if there is a triplet in the given BST with a sum equal to 0, otherwise returns false. 

The expected time complexity should be O(n^2) and only O(Logn) extra space can be used. You can modify the given Binary Search Tree. Note that the height of a Balanced BST is always O(Log n).

For example, isTripletPresent() should return true for following BST because there is a triplet with sum 0, the triplet is {-13, 6, 7}.

The Brute Force Solution is to consider each triplet in BST and check whether the sum adds up to zero. The time complexity of this solution will be O(n^3).

A Better Solution is to create an auxiliary array and store the Inorder traversal of BST in the array. The array will be sorted as Inorder traversal of BST always produces sorted data. Once we have the Inorder traversal, we can use method 2 of this post to find the triplet with a sum equals to 0. This solution works in O(n^2) time but requires O(n) auxiliary space.

Following is the solution that works in O(n^2) time and uses O(Logn) extra space: 

1. Convert given BST to Doubly Linked List (DLL) 
2. Now iterate through every node of DLL and if the key of node is negative, then find a pair in DLL with sum equal to key of current node multiplied by -1. To find the pair, we can use the approach used in hasArrayTwoCandidates() in method 1 of this post. 

Implementation:

C++

// A C++ program to check if there
// is a triplet with sum equal to 0 in
// a given BST
#include <bits/stdc++.h>
using namespace std;

// A BST node has key, and left and right pointers
class node
{
    public:
    int key;
    node *left;
    node *right;
};

// A function to convert given BST to Doubly
// Linked List. left pointer is used
// as previous pointer and right pointer
// is used as next pointer. The function
// sets *head to point to first and *tail
// to point to last node of converted DLL
void convertBSTtoDLL(node* root, node** head, node** tail)
{
    // Base case
    if (root == NULL)
        return;

    // First convert the left subtree
    if (root->left)
        convertBSTtoDLL(root->left, head, tail);

    // Then change left of current root
    // as last node of left subtree
    root->left = *tail;

    // If tail is not NULL, then set right
    // of tail as root, else current
    // node is head
    if (*tail)
        (*tail)->right = root;
    else
        *head = root;

    // Update tail
    *tail = root;

    // Finally, convert right subtree
    if (root->right)
        convertBSTtoDLL(root->right, head, tail);
}

// This function returns true if there
// is pair in DLL with sum equal to given
// sum. The algorithm is similar to hasArrayTwoCandidates()
// in method 1 of http://tinyurl.com/dy6palr
bool isPresentInDLL(node* head, node* tail, int sum)
{
    while (head != tail)
    {
        int curr = head->key + tail->key;
        if (curr == sum)
            return true;
        else if (curr > sum)
            tail = tail->left;
        else
            head = head->right;
    }
    return false;
}

// The main function that returns
// true if there is a 0 sum triplet in
// BST otherwise returns false
bool isTripletPresent(node *root)
{
    // Check if the given BST is empty
    if (root == NULL)
    return false;

    // Convert given BST to doubly linked list. head and tail store the
    // pointers to first and last nodes in DLLL
    node* head = NULL;
    node* tail = NULL;
    convertBSTtoDLL(root, &head, &tail);

    // Now iterate through every node and
    // find if there is a pair with sum
    // equal to -1 * head->key where head is current node
    while ((head->right != tail) && (head->key < 0))
    {
        // If there is a pair with sum
        // equal to -1*head->key, then return
        // true else move forward
        if (isPresentInDLL(head->right, tail, -1*head->key))
            return true;
        else
            head = head->right;
    }

    // If we reach here, then
    // there was no 0 sum triplet
    return false;
}

// A utility function to create
// a new BST node with key as given num
node* newNode(int num)
{
    node* temp = new node();
    temp->key = num;
    temp->left = temp->right = NULL;
    return temp;
}

// A utility function to insert a given key to BST
node* insert(node* root, int key)
{
    if (root == NULL)
    return newNode(key);
    if (root->key > key)
        root->left = insert(root->left, key);
    else
        root->right = insert(root->right, key);
    return root;
}

// Driver code
int main()
{
    node* root = NULL;
    root = insert(root, 6);
    root = insert(root, -13);
    root = insert(root, 14);
    root = insert(root, -8);
    root = insert(root, 15);
    root = insert(root, 13);
    root = insert(root, 7);
    if (isTripletPresent(root))
        cout << "Present";
    else
        cout << "Not Present";
    return 0;
}

// This code is contributed by rathbhupendra

C

// A C program to check if there is a triplet with sum equal to 0 in
// a given BST
#include<stdio.h>

// A BST node has key, and left and right pointers
struct node
{
    int key;
    struct node *left;
    struct node *right;
};

// A function to convert given BST to Doubly Linked List. left pointer is used
// as previous pointer and right pointer is used as next pointer. The function
// sets *head to point to first and *tail to point to last node of converted DLL
void convertBSTtoDLL(node* root, node** head, node** tail)
{
    // Base case
    if (root == NULL)
        return;

    // First convert the left subtree
    if (root->left)
        convertBSTtoDLL(root->left, head, tail);

    // Then change left of current root as last node of left subtree
    root->left = *tail;

    // If tail is not NULL, then set right of tail as root, else current
    // node is head
    if (*tail)
        (*tail)->right = root;
    else
        *head = root;

    // Update tail
    *tail = root;

    // Finally, convert right subtree
    if (root->right)
        convertBSTtoDLL(root->right, head, tail);
}

// This function returns true if there is pair in DLL with sum equal
// to given sum. The algorithm is similar to hasArrayTwoCandidates()
// in method 1 of http://tinyurl.com/dy6palr
bool isPresentInDLL(node* head, node* tail, int sum)
{
    while (head != tail)
    {
        int curr = head->key + tail->key;
        if (curr == sum)
            return true;
        else if (curr > sum)
            tail = tail->left;
        else
            head = head->right;
    }
    return false;
}

// The main function that returns true if there is a 0 sum triplet in
// BST otherwise returns false
bool isTripletPresent(node *root)
{
    // Check if the given  BST is empty
    if (root == NULL)
       return false;

    // Convert given BST to doubly linked list.  head and tail store the
    // pointers to first and last nodes in DLLL
    node* head = NULL;
    node* tail = NULL;
    convertBSTtoDLL(root, &head, &tail);

    // Now iterate through every node and find if there is a pair with sum
    // equal to -1 * head->key where head is current node
    while ((head->right != tail) && (head->key < 0))
    {
        // If there is a pair with sum equal to  -1*head->key, then return
        // true else move forward
        if (isPresentInDLL(head->right, tail, -1*head->key))
            return true;
        else
            head = head->right;
    }

    // If we reach here, then there was no 0 sum triplet
    return false;
}

// A utility function to create a new BST node with key as given num
node* newNode(int num)
{
    node* temp = new node;
    temp->key = num;
    temp->left = temp->right = NULL;
    return temp;
}

// A utility function to insert a given key to BST
node* insert(node* root, int key)
{
    if (root == NULL)
       return newNode(key);
    if (root->key > key)
       root->left = insert(root->left, key);
    else
       root->right = insert(root->right, key);
    return root;
}

// Driver program to test above functions
int main()
{
    node* root = NULL;
    root = insert(root, 6);
    root = insert(root, -13);
    root = insert(root, 14);
    root = insert(root, -8);
    root = insert(root, 15);
    root = insert(root, 13);
    root = insert(root, 7);
    if (isTripletPresent(root))
        printf("Present");
    else
        printf("Not Present");
    return 0;
}

Java

// A Java program to check if there
// is a triplet with sum equal to 0 in
// a given BST
import java.util.*;
class GFG
{

  // A BST node has key, and left and right pointers
  static class node
  {
    int key;
    node left;
    node right;
  };
  static node head;
  static node tail;

  // A function to convert given BST to Doubly
  // Linked List. left pointer is used
  // as previous pointer and right pointer
  // is used as next pointer. The function
  // sets *head to point to first and *tail
  // to point to last node of converted DLL
  static void convertBSTtoDLL(node root)
  {

    // Base case
    if (root == null)
      return;

    // First convert the left subtree
    if (root.left != null)
      convertBSTtoDLL(root.left);

    // Then change left of current root
    // as last node of left subtree
    root.left = tail;

    // If tail is not null, then set right
    // of tail as root, else current
    // node is head
    if (tail != null)
      (tail).right = root;
    else
      head = root;

    // Update tail
    tail = root;

    // Finally, convert right subtree
    if (root.right != null)
      convertBSTtoDLL(root.right);
  }

  // This function returns true if there
  // is pair in DLL with sum equal to given
  // sum. The algorithm is similar to hasArrayTwoCandidates()
  // in method 1 of http://tinyurl.com/dy6palr
  static boolean isPresentInDLL(node head, node tail, int sum)
  {
    while (head != tail)
    {
      int curr = head.key + tail.key;
      if (curr == sum)
        return true;
      else if (curr > sum)
        tail = tail.left;
      else
        head = head.right;
    }
    return false;
  }

  // The main function that returns
  // true if there is a 0 sum triplet in
  // BST otherwise returns false
  static boolean isTripletPresent(node root)
  {
    // Check if the given BST is empty
    if (root == null)
      return false;

    // Convert given BST to doubly linked list. head and tail store the
    // pointers to first and last nodes in DLLL
    head = null;
    tail = null;
    convertBSTtoDLL(root);

    // Now iterate through every node and
    // find if there is a pair with sum
    // equal to -1 * head.key where head is current node
    while ((head.right != tail) && (head.key < 0))
    {
      // If there is a pair with sum
      // equal to -1*head.key, then return
      // true else move forward
      if (isPresentInDLL(head.right, tail, -1*head.key))
        return true;
      else
        head = head.right;
    }

    // If we reach here, then
    // there was no 0 sum triplet
    return false;
  }

  // A utility function to create
  // a new BST node with key as given num
  static node newNode(int num)
  {
    node temp = new node();
    temp.key = num;
    temp.left = temp.right = null;
    return temp;
  }

  // A utility function to insert a given key to BST
  static node insert(node root, int key)
  {
    if (root == null)
      return newNode(key);
    if (root.key > key)
      root.left = insert(root.left, key);
    else
      root.right = insert(root.right, key);
    return root;
  }

  // Driver code
  public static void main(String[] args)
  {
    node root = null;
    root = insert(root, 6);
    root = insert(root, -13);
    root = insert(root, 14);
    root = insert(root, -8);
    root = insert(root, 15);
    root = insert(root, 13);
    root = insert(root, 7);
    if (isTripletPresent(root))
      System.out.print("Present");
    else
      System.out.print("Not Present");
  }
}

// This code is contributed by aashish1995

Python3

# A Python program to check if there
# is a triplet with sum equal to 0 in
# a given BST

# A BST Node has key, and left and right pointers
class Node:
    def __init__(self):
        self.key = 0;
        self.left = None;
        self.right = None;
head = Node();
tail = Node();

# A function to convert given BST to Doubly
# Linked List. left pointer is used
# as previous pointer and right pointer
# is used as next pointer. The function
# sets *head to point to first and *tail
# to point to last Node of converted DLL
def convertBSTtoDLL(root):

    # Base case
    if (root == None):
        return;
    global tail;
    global head;

    # First convert the left subtree
    if (root.left != None):
        convertBSTtoDLL(root.left);

    # Then change left of current root
    # as last Node of left subtree
    root.left = tail;

    # If tail is not None, then set right
    # of tail as root, else current
    # Node is head
    if (tail != None):
        (tail).right = root;
    else:
        head = root;

    # Update tail
    tail = root;

    # Finally, convert right subtree
    if (root.right != None):
        convertBSTtoDLL(root.right);

# This function returns True if there
# is pair in DLL with sum equal to given
# sum. The algorithm is similar to hasArrayTwoCandidates()
# in method 1 of http:#tinyurl.com/dy6palr
def isPresentInDLL(head, tail, s):
    while (head != tail):
        curr = head.key + tail.key;
        if (curr == s):
            return True;
        elif(curr > s):
            tail = tail.left;
        else:
            head = head.right;

    return False;

# The main function that returns
# True if there is a 0 sum triplet in
# BST otherwise returns False
def isTripletPresent(root):

    # Check if the given BST is empty
    if (root == None):
        return False;

    # Convert given BST to doubly linked list. head and tail store the
    # pointers to first and last Nodes in DLLL
    global tail;
    global head;
    head = None;
    tail = None;
    convertBSTtoDLL(root);

    # Now iterate through every Node and
    # find if there is a pair with sum
    # equal to -1 * head.key where head is current Node
    while ((head.right != tail) and (head.key < 0)):

        # If there is a pair with sum
        # equal to -1*head.key, then return
        # True else move forward
        if (isPresentInDLL(head.right, tail, -1 * head.key)):
            return True;
        else:
            head = head.right;


    # If we reach here, then
    # there was no 0 sum triplet
    return False;

# A utility function to create
# a new BST Node with key as given num
def newNode(num):
    temp = Node();
    temp.key = num;
    temp.left = temp.right = None;
    return temp;

# A utility function to insert a given key to BST
def insert(root, key):
    if (root == None):
        return newNode(key);
    if (root.key > key):
        root.left = insert(root.left, key);
    else:
        root.right = insert(root.right, key);
    return root;

# Driver code
if __name__ == '__main__':
    root = None;
    root = insert(root, 6);
    root = insert(root, -13);
    root = insert(root, 14);
    root = insert(root, -8);
    root = insert(root, 15);
    root = insert(root, 13);
    root = insert(root, 7);
    if (isTripletPresent(root)):
        print("Present");
    else:
        print("Not Present");

# This code is contributed by Rajput-Ji

C#

// A C# program to check if there
// is a triplet with sum equal to 0 in
// a given BST
using System;
public class GFG
{

  // A BST node has key, and left and right pointers
  public class node
  {
    public int key;
    public  node left;
    public  node right;
  };
  static node head;
  static node tail;

  // A function to convert given BST to Doubly
  // Linked List. left pointer is used
  // as previous pointer and right pointer
  // is used as next pointer. The function
  // sets *head to point to first and *tail
  // to point to last node of converted DLL
  static void convertBSTtoDLL(node root)
  {

    // Base case
    if (root == null)
      return;

    // First convert the left subtree
    if (root.left != null)
      convertBSTtoDLL(root.left);

    // Then change left of current root
    // as last node of left subtree
    root.left = tail;

    // If tail is not null, then set right
    // of tail as root, else current
    // node is head
    if (tail != null)
      (tail).right = root;
    else
      head = root;

    // Update tail
    tail = root;

    // Finally, convert right subtree
    if (root.right != null)
      convertBSTtoDLL(root.right);
  }

  // This function returns true if there
  // is pair in DLL with sum equal to given
  // sum. The algorithm is similar to hasArrayTwoCandidates()
  // in method 1 of http://tinyurl.com/dy6palr
  static bool isPresentInDLL(node head, node tail, int sum)
  {
    while (head != tail)
    {
      int curr = head.key + tail.key;
      if (curr == sum)
        return true;
      else if (curr > sum)
        tail = tail.left;
      else
        head = head.right;
    }
    return false;
  }

  // The main function that returns
  // true if there is a 0 sum triplet in
  // BST otherwise returns false
  static bool isTripletPresent(node root)
  {

    // Check if the given BST is empty
    if (root == null)
      return false;

    // Convert given BST to doubly linked list. head and tail store the
    // pointers to first and last nodes in DLLL
    head = null;
    tail = null;
    convertBSTtoDLL(root);

    // Now iterate through every node and
    // find if there is a pair with sum
    // equal to -1 * head.key where head is current node
    while ((head.right != tail) && (head.key < 0))
    {
      // If there is a pair with sum
      // equal to -1*head.key, then return
      // true else move forward
      if (isPresentInDLL(head.right, tail, -1*head.key))
        return true;
      else
        head = head.right;
    }

    // If we reach here, then
    // there was no 0 sum triplet
    return false;
  }

  // A utility function to create
  // a new BST node with key as given num
  static node newNode(int num)
  {
    node temp = new node();
    temp.key = num;
    temp.left = temp.right = null;
    return temp;
  }

  // A utility function to insert a given key to BST
  static node insert(node root, int key)
  {
    if (root == null)
      return newNode(key);
    if (root.key > key)
      root.left = insert(root.left, key);
    else
      root.right = insert(root.right, key);
    return root;
  }

  // Driver code
  public static void Main(String[] args)
  {
    node root = null;
    root = insert(root, 6);
    root = insert(root, -13);
    root = insert(root, 14);
    root = insert(root, -8);
    root = insert(root, 15);
    root = insert(root, 13);
    root = insert(root, 7);
    if (isTripletPresent(root))
      Console.Write("Present");
    else
      Console.Write("Not Present");
  }
}

// This code is contributed by aashish1995

Javascript

<script>

// A JavaScript program to check if there
// is a triplet with sum equal to 0 in
// a given BST

    // A BST node has key,
    // and left and right pointers
     class node {
        constructor(val) {
            this.key = val;
            this.left = null;
            this.right = null;
        }
    }

     var head;
     var tail;

    // A function to convert given BST to Doubly
    // Linked List. left pointer is used
    // as previous pointer and right pointer
    // is used as next pointer. The function
    // sets *head to point to first and *tail
    // to point to last node of converted DLL
    function convertBSTtoDLL( root) {

        // Base case
        if (root == null)
            return;

        // First convert the left subtree
        if (root.left != null)
            convertBSTtoDLL(root.left);

        // Then change left of current root
        // as last node of left subtree
        root.left = tail;

        // If tail is not null, then set right
        // of tail as root, else current
        // node is head
        if (tail != null)
            (tail).right = root;
        else
            head = root;

        // Update tail
        tail = root;

        // Finally, convert right subtree
        if (root.right != null)
            convertBSTtoDLL(root.right);
    }

    // This function returns true if there
    // is pair in DLL with sum equal to given
    // sum. The algorithm is
    // similar to hasArrayTwoCandidates()
    // in method 1 of http://tinyurl.com/dy6palr
    function isPresentInDLL( head,  tail , sum) {
        while (head != tail) {
            var curr = head.key + tail.key;
            if (curr == sum)
                return true;
            else if (curr > sum)
                tail = tail.left;
            else
                head = head.right;
        }
        return false;
    }

    // The main function that returns
    // true if there is a 0 sum triplet in
    // BST otherwise returns false
    function isTripletPresent( root) {
        // Check if the given BST is empty
        if (root == null)
            return false;

        // Convert given BST to doubly
        // linked list. head and tail store the
        // pointers to first and last nodes in DLLL
        head = null;
        tail = null;
        convertBSTtoDLL(root);

        // Now iterate through every node and
        // find if there is a pair with sum
        // equal to -1 * head.key where
        // head is current node
        while ((head.right != tail) && (head.key < 0))
        {
            // If there is a pair with sum
            // equal to -1*head.key, then return
            // true else move forward
            if (isPresentInDLL(head.right, tail,
            -1 * head.key))
                return true;
            else
                head = head.right;
        }

        // If we reach here, then
        // there was no 0 sum triplet
        return false;
    }

    // A utility function to create
    // a new BST node with key as given num
     function newNode(num) {
        var temp = new node();
        temp.key = num;
        temp.left = temp.right = null;
        return temp;
    }

    // A utility function to insert a given key to BST
     function insert( root , key) {
        if (root == null)
            return newNode(key);
        if (root.key > key)
            root.left = insert(root.left, key);
        else
            root.right = insert(root.right, key);
        return root;
    }

    // Driver code

        var root = null;
        root = insert(root, 6);
        root = insert(root, -13);
        root = insert(root, 14);
        root = insert(root, -8);
        root = insert(root, 15);
        root = insert(root, 13);
        root = insert(root, 7);
        if (isTripletPresent(root))
            document.write("Present");
        else
            document.write("Not Present");

// This code contributed by gauravrajput1

</script>

Output

Present

Note that the above solution modifies given BST.

Time Complexity: Time taken to convert BST to DLL is O(n) and time taken to find triplet in DLL is O(n^2).
Auxiliary Space: The auxiliary space is needed only for function call stack in recursive function convertBSTtoDLL(). Since given tree is balanced (height is O(Logn)), the number of functions in call stack will never be more than O(Logn).

We can also find triplet in same time and extra space without modifying the tree. See next post. The code discussed there can be used to find triplet also.