538. Convert BST to Greater Tree

Problem

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 [0, 10^4].

• -10^4 <= Node.val <= 10^4

• All the values in the tree are unique.

• root is guaranteed to be a valid binary search tree.

  Note: This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

Solution (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode convertBST(TreeNode root) {
        if (root != null) {
            postOrder(root, 0);
        }
        return root;
    }

    private int postOrder(TreeNode node, int val) {
        int newVal = 0;
        if (node.right != null) {
            newVal += postOrder(node.right, val);
        }
        newVal += newVal == 0 ? val + node.val : node.val;
        node.val = newVal;
        if (node.left != null) {
            newVal = postOrder(node.left, newVal);
        }
        return newVal;
    }
}

Explain:

nope.

Complexity:

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