LeetCode in Kotlin
1038. Binary Search Tree to Greater Sum 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
[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/
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 {
private var greaterSum = 0
fun bstToGst(root: TreeNode?): TreeNode {
if (root!!.right != null) {
bstToGst(root.right!!)
}
root.`val` = greaterSum + root.`val`
greaterSum = root.`val`
if (root.left != null) {
bstToGst(root.left!!)
}
return root
}
}