0538. Convert BST to Greater Tree (M)

Convert BST to Greater Tree (M)

题目

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 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.

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

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]

Example 3:

Input: root = [1,0,2]
Output: [3,3,2]

Example 4:

Input: root = [3,2,4,1]
Output: [7,9,4,10]

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.

题意

对一个二叉搜索树T中的结点进行修改，使每个结点的值变为原值x加上T中所有大于x的值的和。

思路

由于BST的中序遍历序列S是递增，每个结点的值实际上变成了原值x加上S中所有排在x后面的值，因此可以从S中最后一个结点开始向前处理，逐步累加值即可，做一次"逆"中序遍历。

代码实现

Java

class Solution {
    private int sum = 0;

    public TreeNode convertBST(TreeNode root) {
        if (root == null) return null;

        convertBST(root.right);
        sum += root.val;
        root.val = sum;
        convertBST(root.left);

        return root;
    }
}