1038. Binary Search Tree to Greater Sum Tree

Description

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/

 

Solutions

Solution: Depth-First Search

• Time complexity: O(n)
• Space complexity: O(logn)

 

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
const bstToGst = function (root) {
  let sum = 0;

  const sumBst = node => {
    if (!node) return;
    const { val, left, right } = node;

    sumBst(right);
    sum += val;
    node.val = sum;
    sumBst(left);
  };

  sumBst(root);
  return root;
};