Pair with a given sum in BST | Set 2

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:

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// 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;
}

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Output:

Yes


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