Algorithm

给定一个二叉树，找出其最大深度。

二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。

说明: 叶子节点是指没有子节点的节点。

示例： 给定二叉树 [3,9,20,null,null,15,7]，

3复制代码

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/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public int maxDepth(TreeNode root) {        if(root == null){            return 0;        }                int number = 1 ;        int leftDepth = 0,rightDepth = 0;        if(root.left != null){            leftDepth = maxDepth(root.left);        }                if(root.right != null){            rightDepth = maxDepth(root.right);        }        number += leftDepth > rightDepth ? leftDepth : rightDepth;        return number;    }}复制代码

Review

The gender gap is also a confidence gap

Tip

之前有个朋友在写sql的时候碰到一个问题写sql查询的时候碰到了问题：

#sql1select  u.real_name,u.phone,b.order_no,b.amount as borrow_amount,b.create_time as borrow_time,r.id,r.user_id,r.borrow_id,r.state,r.amount  as repay_amount,r.repay_time,r.penalty_amout,r.penalty_day,         b.fee ,b.real_amount,channel.name as channel_name,cmro.state as allotState,u.living_img,u.front_img,u.back_img, r.remark        from  cl_borrow_repay r  left join  cl_user_base_info u on  u.user_id=r.user_id  join cl_borrow b on r.borrow_id=b.id         left join cl_user user2 on user2.id = u.user_id         left join cl_channel channel on user2.channel_id = channel.id         left join cl_manual_repay_order cmro on r.id = cmro.borrow_repay_id;#sql2select  u.real_name,u.phone,b.order_no,b.amount as borrow_amount,b.create_time as borrow_time,r.id,r.user_id,r.borrow_id,r.state,r.amount  as repay_amount,r.repay_time,r.penalty_amout,r.penalty_day,         b.fee ,b.real_amount,channel.name as channel_name,cmro.state as allotState,u.living_img,u.front_img,u.back_img, r.remark        from  cl_borrow_repay r  left join  cl_user_base_info u on  u.user_id=r.user_id  join cl_borrow b on r.borrow_id=b.id         left join cl_user user2 on user2.id = u.user_id         left join cl_channel channel on user2.channel_id = channel.id         left join cl_manual_repay_order cmro on r.id = cmro.borrow_repay_idORDER BY r.id desc;复制代码

sql1与sql2上基本（是不是应该用专业点的属于表示）没有改动，仅仅是在末位加上了order by ，但是最后的查询时间有着天壤之别。explain之后的结果如下 sql1:

发现在Extra中多了Using temporary；Using filesort。 原来查询慢的原因是因为使用orderby 后导致sql查询新增加了临时表及进行了文件排序。大量时间花在这上面了。 那么怎么优化？ 1、可以考虑在条件中把主键id加入进去左右一起的条件，因为如果order by 的字段也在where里面这样就不会进入文件排序 2、可以参考一次mysql 优化 （Using temporary ; Using filesort）

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JVM发生频繁 CMS GC，罪魁祸首是这个参数！ : .