Skip to content
Related Articles

Related Articles

Improve Article
Pair with a given sum in BST | Set 2
  • Difficulty Level : Medium
  • Last Updated : 07 Jun, 2021

Given a binary search tree, and an integer X, the task is to check if there exists a pair of distinct nodes in BST with sum equal to X. If yes then print Yes else print No.

Examples: 

Input: X = 5
          5 
        /   \ 
       3     7 
      / \   / \ 
     2   4 6   8
Output: Yes
2 + 3 = 5. Thus, the answer is "Yes"

Input: X = 10
      1
       \
        2
         \
          3
           \
            4
             \
              5
Output: No

Approach: We have already discussed a hash based approach in this article. The space complexity of this is O(N) where N is the number of nodes in BST.

In this article, we will solve the same problem using a space efficient method by reducing the space complexity to O(H) where H is the height of BST. For that, we will use two pointer technique on BST. Thus, we will maintain a forward and a backward iterator that will iterate the BST in the order of in-order and reverse in-order traversal respectively. Following are the steps to solve the problem: 

  1. Create a forward and backward iterator for BST. Let’s say the value of nodes they are pointing at are v1 and v2.
  2. Now at each step, 
    • If v1 + v2 = X, we found a pair.
    • If v1 + v2 < x, we will make forward iterator point to the next element.
    • If v1 + v2 > x, we will make backward iterator point to the previous element.
  3. If we find no such pair, answer will be “No”.

Below is the implementation of the above approach: 



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Node of the binary tree
struct node {
    int data;
    node* left;
    node* right;
    node(int data)
    {
        this->data = data;
        left = NULL;
        right = NULL;
    }
};
 
// Function to find a pair with given sum
bool existsPair(node* root, int x)
{
    // Iterators for BST
    stack<node *> it1, it2;
 
    // Initializing forward iterator
    node* c = root;
    while (c != NULL)
        it1.push(c), c = c->left;
 
    // Initializing backward iterator
    c = root;
    while (c != NULL)
        it2.push(c), c = c->right;
 
    // Two pointer technique
    while (it1.top() != it2.top()) {
 
        // Variables to store values at
        // it1 and it2
        int v1 = it1.top()->data, v2 = it2.top()->data;
 
        // Base case
        if (v1 + v2 == x)
            return true;
 
        // Moving forward pointer
        if (v1 + v2 < x) {
            c = it1.top()->right;
            it1.pop();
            while (c != NULL)
                it1.push(c), c = c->left;
        }
 
        // Moving backward pointer
        else {
            c = it2.top()->left;
            it2.pop();
            while (c != NULL)
                it2.push(c), c = c->right;
        }
    }
 
    // Case when no pair is found
    return false;
}
 
// Driver code
int main()
{
    node* root = new node(5);
    root->left = new node(3);
    root->right = new node(7);
    root->left->left = new node(2);
    root->left->right = new node(4);
    root->right->left = new node(6);
    root->right->right = new node(8);
 
    int x = 5;
 
    // Calling required function
    if (existsPair(root, x))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Node of the binary tree
static class node
{
    int data;
    node left;
    node right;
    node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
};
 
// Function to find a pair with given sum
static boolean existsPair(node root, int x)
{
    // Iterators for BST
    Stack<node > it1 = new Stack<node>(), it2 = new Stack<node>();
 
    // Initializing forward iterator
    node c = root;
    while (c != null)
    {
        it1.push(c);
        c = c.left;
    }
 
    // Initializing backward iterator
    c = root;
    while (c != null)
    {
        it2.push(c);
        c = c.right;
    }
         
    // Two pointer technique
    while (it1.peek() != it2.peek())
    {
 
        // Variables to store values at
        // it1 and it2
        int v1 = it1.peek().data, v2 = it2.peek().data;
 
        // Base case
        if (v1 + v2 == x)
            return true;
 
        // Moving forward pointer
        if (v1 + v2 < x)
        {
            c = it1.peek().right;
            it1.pop();
            while (c != null)
            {
                it1.push(c);
                c = c.left;
            }
        }
 
        // Moving backward pointer
        else
        {
            c = it2.peek().left;
            it2.pop();
            while (c != null)
            {
                it2.push(c);
                c = c.right;
            }
        }
    }
     
    // Case when no pair is found
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    node root = new node(5);
    root.left = new node(3);
    root.right = new node(7);
    root.left.left = new node(2);
    root.left.right = new node(4);
    root.right.left = new node(6);
    root.right.right = new node(8);
 
    int x = 5;
 
    // Calling required function
    if (existsPair(root, x))
        System.out.print("Yes");
    else
        System.out.print("No");
 
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
 
# Node of the binary tree
class node:
     
    def __init__ (self, key):
         
        self.data = key
        self.left = None
        self.right = None
 
# Function that returns true if a pair
# with given sum exists in the given BSTs
def existsPair(root1, x):
     
    # Stack to store nodes for forward
    # and backward iterator
    it1, it2 = [], []
 
    # Initializing forward iterator
    c = root1
    while (c != None):
        it1.append(c)
        c = c.left
 
    # Initializing backward iterator
    c = root1
    while (c != None):
        it2.append(c)
        c = c.right
 
    # Two pointer technique
    while (it1[-1] != it2[-1]):
 
        # To store the value of the nodes
        # current iterators are pointing to
        v1 = it1[-1].data
        v2 = it2[-1].data
 
        # Base case
        if (v1 + v2 == x):
            return True
 
        # Moving forward iterator
        if (v1 + v2 < x):
            c = it1[-1].right
            del it1[-1]
             
            while (c != None):
                it1.append(c)
                c = c.left
 
        # Moving backward iterator
        else:
            c = it2[-1].left
            del it2[-1]
             
            while (c != None):
                it2.append(c)
                c = c.right
 
    # If no such pair found
    return False
 
# Driver code
if __name__ == '__main__':
 
    root2 = node(5)
    root2.left = node(3)
    root2.right = node(7)
    root2.left.left = node(2)
    root2.left.right = node(4)
    root2.right.left = node(6)
    root2.right.right = node(8)
 
    x = 5
     
    # Calling required function
    if (existsPair(root2, x)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Node of the binary tree
public class node
{
    public int data;
    public node left;
    public node right;
    public node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
};
 
// Function to find a pair with given sum
static bool existsPair(node root, int x)
{
    // Iterators for BST
    Stack<node > it1 = new Stack<node>(),
                 it2 = new Stack<node>();
 
    // Initializing forward iterator
    node c = root;
    while (c != null)
    {
        it1.Push(c);
        c = c.left;
    }
 
    // Initializing backward iterator
    c = root;
    while (c != null)
    {
        it2.Push(c);
        c = c.right;
    }
         
    // Two pointer technique
    while (it1.Peek() != it2.Peek())
    {
 
        // Variables to store values at
        // it1 and it2
        int v1 = it1.Peek().data,
            v2 = it2.Peek().data;
 
        // Base case
        if (v1 + v2 == x)
            return true;
 
        // Moving forward pointer
        if (v1 + v2 < x)
        {
            c = it1.Peek().right;
            it1.Pop();
            while (c != null)
            {
                it1.Push(c);
                c = c.left;
            }
        }
 
        // Moving backward pointer
        else
        {
            c = it2.Peek().left;
            it2.Pop();
            while (c != null)
            {
                it2.Push(c);
                c = c.right;
            }
        }
    }
     
    // Case when no pair is found
    return false;
}
 
// Driver code
public static void Main(String[] args)
{
    node root = new node(5);
    root.left = new node(3);
    root.right = new node(7);
    root.left.left = new node(2);
    root.left.right = new node(4);
    root.right.left = new node(6);
    root.right.right = new node(8);
 
    int x = 5;
 
    // Calling required function
    if (existsPair(root, x))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
// Node of the binary tree
class node
{
    constructor(data)
    {
        this.data = data;
        this.left = this.right = null;
    }
}
 
// Function to find a pair with given sum
function existsPair(root, x)
{
     
    // Iterators for BST
    let it1 = [], it2 = [];
  
    // Initializing forward iterator
    let c = root;
    while (c != null)
    {
        it1.push(c);
        c = c.left;
    }
  
    // Initializing backward iterator
    c = root;
    while (c != null)
    {
        it2.push(c);
        c = c.right;
    }
          
    // Two pointer technique
    while (it1[it1.length-1] != it2[it2.length-1])
    {
  
        // Variables to store values at
        // it1 and it2
        let v1 = it1[it1.length - 1].data,
            v2 = it2[it2.length - 1].data;
  
        // Base case
        if (v1 + v2 == x)
            return true;
  
        // Moving forward pointer
        if (v1 + v2 < x)
        {
            c = it1[it1.length - 1].right;
            it1.pop();
             
            while (c != null)
            {
                it1.push(c);
                c = c.left;
            }
        }
  
        // Moving backward pointer
        else
        {
            c = it2[it2.length - 1].left;
            it2.pop();
             
            while (c != null)
            {
                it2.push(c);
                c = c.right;
            }
        }
    }
      
    // Case when no pair is found
    return false;
}
 
// Driver code
let root = new node(5);
root.left = new node(3);
root.right = new node(7);
root.left.left = new node(2);
root.left.right = new node(4);
root.right.left = new node(6);
root.right.right = new node(8);
 
let x = 5;
 
// Calling required function
if (existsPair(root, x))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by unknown2108
 
</script>
Output: 
Yes

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live 




My Personal Notes arrow_drop_up
Recommended Articles
Page :