1038. Binary Search Tree to Greater Sum Tree LeetCode Solution

In this guide, you will get 1038. Binary Search Tree to Greater Sum Tree LeetCode Solution with the best time and space complexity. The solution to Binary Search Tree to Greater Sum Tree problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Binary Search Tree to Greater Sum Tree solution in C++
Binary Search Tree to Greater Sum Tree solution in Java
Binary Search Tree to Greater Sum Tree solution in Python
Additional Resources

1038. Binary Search Tree to Greater Sum Tree LeetCode Solution image

Problem Statement of Binary Search Tree to Greater Sum Tree

Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.
As a reminder, a binary search tree is a tree that satisfies these constraints:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than the node’s key.
Both the left and right subtrees must also be binary search trees.

Example 1:

Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Example 2:

Input: root = [0,null,1]
Output: [1,null,1]

Constraints:

The number of nodes in the tree is in the range [1, 100].
0 <= Node.val <= 100
All the values in the tree are unique.

Note: This question is the same as 538: https://leetcode.com/problems/convert-bst-to-greater-tree/

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(\log n)

1038. Binary Search Tree to Greater Sum Tree LeetCode Solution in C++

class Solution {
 public:
  TreeNode* bstToGst(TreeNode* root) {
    int prefix = 0;

    function<void(TreeNode*)> reversedInorder = [&](TreeNode* root) {
      if (root == nullptr)
        return;

      reversedInorder(root->right);

      root->val += prefix;
      prefix = root->val;

      reversedInorder(root->left);
    };

    reversedInorder(root);
    return root;
  }
};
/* code provided by PROGIEZ */