LeetCode in Kotlin

538. Convert BST to Greater Tree

Medium

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, 104].
• -104 <= Node.val <= 104
• 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

import com_github_leetcode.TreeNode

/*
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun convertBST(root: TreeNode?): TreeNode? {
        if (root != null) {
            postOrder(root, 0)
        }
        return root
    }

    private fun postOrder(node: TreeNode, `val`: Int): Int {
        var newVal = 0
        if (node.right != null) {
            newVal += postOrder(node.right!!, `val`)
        }
        newVal += if (newVal == 0) `val` + node.`val` else node.`val`
        node.`val` = newVal
        if (node.left != null) {
            newVal = postOrder(node.left!!, newVal)
        }
        return newVal
    }
}

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