# Nodes from given two BSTs with sum equal to X

Given two Binary search trees and an integer X, the task is to find a pair of nodes, one belonging to the first BST and the second belonging to other such that their sum is equal to X. If there exists such a pair, print Yes else print No.

Examples:

```Input: X = 100
BST 1:
5
/   \
3     7
/ \   / \
2   4 6   8
BST 2:
11
\
13
Output: No
There is no such pair with given value.

Input: X = 16
BST 1:
5
/   \
3     7
/ \   / \
2   4 6   8
BST 2:
11
\
13
Output: Yes
5 + 11 = 16
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We will solve this problem using two pointer approach.
We will create a forward iterator on the first BST and backward on the second. Thus, we will maintain forward and a backward iterator that will iterate the BSTs in the order of in-order and reverse in-order traversal respectively.

1. Create a forward and backward iterator for first and second BST respectively. 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 less than or equal to x, we will make forward iterator point to the next element.
• If v1 + v2 greater than x, we will make backward iterator point to the previous element.
3. We will continue the above while both iterators are pointing to a valid node.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `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 that returns true if a pair ` `// with given sum exists in the given BSTs ` `bool` `existsPair(node* root1, node* root2, ``int` `x) ` `{ ` `    ``// Stack to store nodes for forward and backward ` `    ``// iterator ` `    ``stack it1, it2; ` ` `  `    ``// Initializing forward iterator ` `    ``node* c = root1; ` `    ``while` `(c != NULL) ` `        ``it1.push(c), c = c->left; ` ` `  `    ``// Initializing backward iterator ` `    ``c = root2; ` `    ``while` `(c != NULL) ` `        ``it2.push(c), c = c->right; ` ` `  `    ``// Two pointer technique ` `    ``while` `(it1.size() and it2.size()) { ` ` `  `        ``// To store the value of the nodes ` `        ``// current iterators are pointing to ` `        ``int` `v1 = it1.top()->data, v2 = it2.top()->data; ` ` `  `        ``// If found a valid pair ` `        ``if` `(v1 + v2 == x) ` `            ``return` `true``; ` ` `  `        ``// Moving forward iterator ` `        ``if` `(v1 + v2 < x) { ` `            ``c = it1.top()->right; ` `            ``it1.pop(); ` `            ``while` `(c != NULL) ` `                ``it1.push(c), c = c->left; ` `        ``} ` ` `  `        ``// Moving backward iterator ` `        ``else` `{ ` `            ``c = it2.top()->left; ` `            ``it2.pop(); ` `            ``while` `(c != NULL) ` `                ``it2.push(c), c = c->right; ` `        ``} ` `    ``} ` ` `  `    ``// If no such pair found ` `    ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``// First BST ` `    ``node* root1 = ``new` `node(11); ` `    ``root1->right = ``new` `node(15); ` ` `  `    ``// Second BST ` `    ``node* root2 = ``new` `node(5); ` `    ``root2->left = ``new` `node(3); ` `    ``root2->right = ``new` `node(7); ` `    ``root2->left->left = ``new` `node(2); ` `    ``root2->left->right = ``new` `node(4); ` `    ``root2->right->left = ``new` `node(6); ` `    ``root2->right->right = ``new` `node(8); ` ` `  `    ``int` `x = 23; ` ` `  `    ``if` `(existsPair(root1, root2, 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 that returns true if a pair ` `// with given sum exists in the given BSTs ` `static` `boolean` `existsPair(node root1, node root2, ``int` `x) ` `{ ` `    ``// Stack to store nodes for forward and backward ` `    ``// iterator ` `    ``Stack it1 = ``new` `Stack(), it2 = ``new` `Stack(); ` ` `  `    ``// Initializing forward iterator ` `    ``node c = root1; ` `    ``while` `(c != ``null``) ` `    ``{ ` `        ``it1.push(c); ` `        ``c = c.left; ` `    ``} ` `    ``// Initializing backward iterator ` `    ``c = root2; ` `    ``while` `(c != ``null``) ` `    ``{ ` `        ``it2.push(c); ` `        ``c = c.right; ` `    ``} ` ` `  `    ``// Two pointer technique ` `    ``while` `(it1.size() > ``0` `&& it2.size() > ``0``) ` `    ``{ ` ` `  `        ``// To store the value of the nodes ` `        ``// current iterators are pointing to ` `        ``int` `v1 = it1.peek().data, v2 = it2.peek().data; ` ` `  `        ``// If found a valid pair ` `        ``if` `(v1 + v2 == x) ` `            ``return` `true``; ` ` `  `        ``// Moving forward iterator ` `        ``if` `(v1 + v2 < x)  ` `        ``{ ` `            ``c = it1.peek().right; ` `            ``it1.pop(); ` `            ``while` `(c != ``null``) ` `            ``{ ` `                ``it1.push(c); c = c.left; ` `            ``} ` `        ``} ` ` `  `        ``// Moving backward iterator ` `        ``else` `        ``{ ` `            ``c = it2.peek().left; ` `            ``it2.pop(); ` `            ``while` `(c != ``null``) ` `            ``{ ` `                ``it2.push(c); c = c.right; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If no such pair found ` `    ``return` `false``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``// First BST ` `    ``node root1 = ``new` `node(``11``); ` `    ``root1.right = ``new` `node(``15``); ` ` `  `    ``// Second BST ` `    ``node root2 = ``new` `node(``5``); ` `    ``root2.left = ``new` `node(``3``); ` `    ``root2.right = ``new` `node(``7``); ` `    ``root2.left.left = ``new` `node(``2``); ` `    ``root2.left.right = ``new` `node(``4``); ` `    ``root2.right.left = ``new` `node(``6``); ` `    ``root2.right.right = ``new` `node(``8``); ` ` `  `    ``int` `x = ``23``; ` ` `  `    ``if` `(existsPair(root1, root2, x)) ` `        ``System.out.println(``"Yes"``); ` `    ``else` `        ``System.out.println(``"No"``); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## 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 that returns true if a pair ` `// with given sum exists in the given BSTs ` `static` `bool` `existsPair(node root1, node root2, ``int` `x) ` `{ ` `    ``// Stack to store nodes for forward and backward ` `    ``// iterator ` `    ``Stack it1 = ``new` `Stack(), it2 = ``new` `Stack(); ` ` `  `    ``// Initializing forward iterator ` `    ``node c = root1; ` `    ``while` `(c != ``null``) ` `    ``{ ` `        ``it1.Push(c); ` `        ``c = c.left; ` `    ``} ` `     `  `    ``// Initializing backward iterator ` `    ``c = root2; ` `    ``while` `(c != ``null``) ` `    ``{ ` `        ``it2.Push(c); ` `        ``c = c.right; ` `    ``} ` ` `  `    ``// Two pointer technique ` `    ``while` `(it1.Count > 0 && it2.Count > 0) ` `    ``{ ` ` `  `        ``// To store the value of the nodes ` `        ``// current iterators are pointing to ` `        ``int` `v1 = it1.Peek().data, v2 = it2.Peek().data; ` ` `  `        ``// If found a valid pair ` `        ``if` `(v1 + v2 == x) ` `            ``return` `true``; ` ` `  `        ``// Moving forward iterator ` `        ``if` `(v1 + v2 < x)  ` `        ``{ ` `            ``c = it1.Peek().right; ` `            ``it1.Pop(); ` `            ``while` `(c != ``null``) ` `            ``{ ` `                ``it1.Push(c); c = c.left; ` `            ``} ` `        ``} ` ` `  `        ``// Moving backward iterator ` `        ``else` `        ``{ ` `            ``c = it2.Peek().left; ` `            ``it2.Pop(); ` `            ``while` `(c != ``null``) ` `            ``{ ` `                ``it2.Push(c); c = c.right; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// If no such pair found ` `    ``return` `false``; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``// First BST ` `    ``node root1 = ``new` `node(11); ` `    ``root1.right = ``new` `node(15); ` ` `  `    ``// Second BST ` `    ``node root2 = ``new` `node(5); ` `    ``root2.left = ``new` `node(3); ` `    ``root2.right = ``new` `node(7); ` `    ``root2.left.left = ``new` `node(2); ` `    ``root2.left.right = ``new` `node(4); ` `    ``root2.right.left = ``new` `node(6); ` `    ``root2.right.right = ``new` `node(8); ` ` `  `    ``int` `x = 23; ` ` `  `    ``if` `(existsPair(root1, root2, x)) ` `        ``Console.WriteLine(``"Yes"``); ` `    ``else` `        ``Console.WriteLine(``"No"``); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

Output:

```Yes
```

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