Find if there is a triplet in a Balanced BST that adds to zero in C++

Suppose we have a balanced binary search tree, we have to create a function named is_valid_triplet() that returns true when there exist a triplet in given BST whose sum equals to 0, otherwise returns false. Design the method by following these constraints −

• expected time complexity is O(n^2)

• O(logn) extra space can be used.

So, if the input is like

then the output will be True, as triplet is [-15,7,8]

To solve this, we will follow these steps −

• Define a function bst_to_doubli_list(), this will take root, head, tail,

• if root is same as NULL, then −

  • return

• if left of root is not null, then −

  • bst_to_doubli_list(left of root, head, tail)

• left of root := tail

• if tail is not null , then −

  • right of tail := root

• Otherwise

  • head := root

• tail := root

• if right of root is not null, then −

  • bst_to_doubli_list(right of root, head, tail)

• Define a function is_in_double_list(), this will take head, tail, sum,

• while head is not equal to tail, do −

  • current := key of head + key of tail

  • if current is same as sum, then −

    • return true

  • otherwise when current > sum, then −

    • tail := left of tail

  • Otherwise

    • head := right of head

• return false

• From the main method, do the following −

• if root is null, then −

  • return false

• head = null

• tail = null

• bst_to_doubli_list(root, head, tail)

• while (right of head is not equal to tail and key of head < 0), do −

  • if is_in_double(right of head, tail, key of head * (-1), then

    • return true

  • Otherwise

    • head := right of head

• return false

Example (C++)

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class TreeNode {
   public:
   int key;
   TreeNode *left;
   TreeNode *right;
   TreeNode() : key(0), left(NULL), right(NULL) {}
   TreeNode(int x) : key(x), left(NULL), right(NULL) {}
};
void bst_to_doubli_list(TreeNode* root, TreeNode** head, TreeNode** tail) {
   if (root == NULL)
      return;
   if (root->left)
      bst_to_doubli_list(root->left, head, tail);
   root->left = *tail;
   if (*tail)
      (*tail)->right = root;
   else
      *head = root;
      *tail = root;
   if (root->right)
      bst_to_doubli_list(root->right, head, tail);
}
bool is_in_double_list(TreeNode* head, TreeNode* tail, int sum) {
   while (head != tail) {
      int current = head->key + tail->key;
      if (current == sum)
         return true;
      else if (current > sum)
         tail = tail->left;
      else
         head = head->right;
   }
   return false;
}
bool is_valid_triplet(TreeNode *root) {
   if (root == NULL)
      return false;
   TreeNode* head = NULL;
   TreeNode* tail = NULL;
   bst_to_doubli_list(root, &head, &tail);
   while ((head->right != tail) && (head->key < 0)){
      if (is_in_double_list(head->right, tail, -1*head->key))
         return true;
      else
         head = head->right;
   }
   return false;
}
TreeNode* insert(TreeNode* root, int key) {
   if (root == NULL)
      return new TreeNode(key);
   if (root->key > key)
      root->left = insert(root->left, key);
   else
      root->right = insert(root->right, key);
   return root;
}
int main(){
   TreeNode* root = NULL;
   root = insert(root, 7);
   root = insert(root, -15);
   root = insert(root, 15);
   root = insert(root, -7);
   root = insert(root, 14);
   root = insert(root, 16);
   root = insert(root, 8);
   cout << is_valid_triplet(root);
}

Input

root = insert(root, 7);
root = insert(root, -15);
root = insert(root, 15);
root = insert(root, -7);
root = insert(root, 14);
root = insert(root, 16);
root = insert(root, 8);

Output

1