惯性聚合 高效追踪和阅读你感兴趣的博客、新闻、科技资讯
阅读原文 在惯性聚合中打开

推荐订阅源

Y
Y Combinator Blog
让小产品的独立变现更简单 - ezindie.com
让小产品的独立变现更简单 - ezindie.com
OSCHINA 社区最新新闻
OSCHINA 社区最新新闻
V
V2EX
博客园 - 三生石上(FineUI控件)
Hugging Face - Blog
Hugging Face - Blog
奇客Solidot–传递最新科技情报
奇客Solidot–传递最新科技情报
罗磊的独立博客
博客园_首页
量子位
雷峰网
雷峰网
GbyAI
GbyAI
小众软件
小众软件
酷 壳 – CoolShell
酷 壳 – CoolShell
D
DataBreaches.Net
H
Hackread – Cybersecurity News, Data Breaches, AI and More
The Cloudflare Blog
IT之家
IT之家
WordPress大学
WordPress大学
人人都是产品经理
人人都是产品经理
Apple Machine Learning Research
Apple Machine Learning Research
P
Proofpoint News Feed
钛媒体:引领未来商业与生活新知
钛媒体:引领未来商业与生活新知
博客园 - 聂微东

博客园 - NickyYe

DYNAMIC LINK LIBRARY - DLL 分布式系统中Unique ID 的生成方法 205. Isomorphic Strings 201. Bitwise AND of Numbers Range 189. Rotate Array 187. Repeated DNA Sequences 167. Two Sum II - Input array is sorted Uncommon Words from Two Sentences Path Sum III Delete Node in a BST Sliding Window Maximum Find K Closest Elements C++ TUTORIAL - MEMORY ALLOCATION - 2016 多线程 Console Event Handling SetConsoleCtrlHandler() -- 设置控制台信号处理函数 SetConsoleCtrlHandler 处理控制台消息 总结open与fopen的区别 LevelDB
Convert BST to Greater Tree
NickyYe · 2018-11-21 · via 博客园 - NickyYe

https://leetcode.com/problems/convert-bst-to-greater-tree/

Given 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.

Example:

Input: The root of a Binary Search Tree like this:
              5
            /   \
           2     13

Output: The root of a Greater Tree like this:
             18
            /   \
          20     13

解题思路:

右子树->root->左子树的顺序,从下往上开始处理。

对于每个节点,记录往前的累加和sum,在置到root上。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode convertBST(TreeNode root) {
        int[] sum = new int[1];
        helper(root, sum);
        return root;
    }
    
    public void helper(TreeNode root, int[] sum) {
        if (root ==null) {
            return;
        }
        helper(root.right, sum);        
        root.val += sum[0];
        sum[0] = root.val;
        helper(root.left, sum);
    }
}