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Pair with a given sum in BST | Set 2
  • Difficulty Level : Medium
  • Last Updated : 05 Nov, 2020
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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
Output: 
Yes




 

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