Triplet with a given sum in BST | Set 2

Given a binary search tree, and an integer X, the task is to find if there exists a triplet with sum X. Print Yes or No correspondingly. Note that the three nodes may not necessarily be distinct.

Examples:

Input: X = 15
          5 
        /   \ 
       3     7 
      / \   / \ 
     2   4 6   8
Output: Yes
{5, 5, 5} is one such triplet.
{3, 5, 7}, {2, 5, 8}, {4, 5, 6} are some others.
  
Input: X = 16
      1
       \
        2
         \
          3
           \
            4
             \
              5
Output: No

Simple Approach: A simple approach will be to convert the BST to a sorted array and then find the triplet using three pointers. This will take O(N) extra space where N is the number of nodes present in the Binary Search Tree. We have already discussed a similar problem in this article which takes O(N) extra space.

Better approach: We will solve this problem using a space efficient method by reducing the additional space complexity to O(H) where H is the height of BST. For that, we will use two pointer technique on BST.
We will traverse all the nodes for the tree one by one and for each node, we will try to find a pair with a sum equal to (X – curr->data) where ‘curr’ is the current node of the BST we are traversing.
We will use a technique similar to the technique discussed in this article for finding a pair.

Algorithm: Traverse each node of BST one by one and for each node:

  1. Create a forward and backward iterator for BST. Lets 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, thus we will increase the count by 1.
    • 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 the left iterator doesn’t point to a node with larger value than right node.

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 that returns true if a pair exists
// in the binary search tree with sum equal to x
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.size() and it2.size()) {
  
        // Variables to store values at
        // it1 and it2
        int v1 = it1.top()->data, v2 = it2.top()->data;
  
        // Base case
        if (v1 + v2 == x)
            return 1;
  
        if (v1 > v2)
            break;
  
        // 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 0;
}
  
// Function that returns true if a triplet exists
// in the binary search tree with sum equal to x
bool existsTriplet(node* root, node* curr, int x)
{
    // If current node is NULL
    if (curr == NULL)
        return 0;
  
    // Conditions for existence of a triplet
    return (existsPair(root, x - curr->data)
            || existsTriplet(root, curr->left, x)
            || existsTriplet(root, curr->right, x));
}
  
// 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 = 24;
  
    if (existsTriplet(root, root, x))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

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Output:

Yes

Time complexity: O(N2)
Space complexity: O(H)



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