Algorithm
题目 112. Path Sum
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Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
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基本思路:
- 使用 DFS 遍历所有的叶节点，计算叶节点到根节点路径的节点 val 之和，等于给定数字则返回 true
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| class Solution {
private boolean hasPath = false;
private int sum = 0;
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
this.sum = sum;
this.dfs(root, 0);
return this.hasPath;
}
public void dfs(TreeNode node, int num) {
if (this.hasPath) {
return;
}
num += node.val;
if (node.left == null && node.right == null) {
if (num == this.sum){
this.hasPath = true;
}
}
if (node.left != null ){
this.dfs(node.left, num);
}
if (node.right != null ) {
this.dfs(node.right, num);
}
}
}
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Review
日志的小短文 The Problem With Logging.
全文可以概括为： “No log everything”,”rules and regulations”
- 日志打印过多，影响代码的可读性与可维护性
- 消耗资源
- 日志信息过多导致日后检索日志变得麻烦
- 日志过多，但真正有用的信息很少
Tip
关于测试中的多线程与jvm相关场景思考：
https://cloudfeng.github.io/2018/06/29/arts/tip/T-java-thread-unit/
Share
Redis 数据持久化理解：
Redis 提供了两种方式: 快照与只追加文件。与 MySQL 的 dump 和 binlog 备份原理是基本一致。前者是将已经存储的数据导出进行备份，后者是根据对存储单元的修改操作进行备份。
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