Java [Leetcode 101]Symmetric Tree
题目描述:
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
解题思路:
递归法
代码如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if(root == null)
return true;
return isMirror(root.left, root.right);
} public boolean isMirror(TreeNode left, TreeNode right){
if(left == null && right == null)
return true;
else if(left == null)
return false;
else if(right == null)
return false;
else{
if(left.val != right.val)
return false;
return isMirror(left.right, right.left) && isMirror(left.left, right.right);
}
}
}
Java [Leetcode 101]Symmetric Tree的更多相关文章
- [leetcode] 101. Symmetric Tree 对称树
题目大意 #!/usr/bin/env python # coding=utf-8 # Date: 2018-08-30 """ https://leetcode.com ...
- Leetcode 101 Symmetric Tree 二叉树
判断一棵树是否自对称 可以回忆我们做过的Leetcode 100 Same Tree 二叉树和Leetcode 226 Invert Binary Tree 二叉树 先可以将左子树进行Invert B ...
- Java for LeetCode 101 Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
- leetcode 101 Symmetric Tree ----- java
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
- LeetCode 101. Symmetric Tree (对称树)
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
- (二叉树 DFS 递归) leetcode 101. Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
- LeetCode 101. Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
- LeetCode 101. Symmetric Tree 判断对称树 C++
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
- Leetcode 101. Symmetric Tree(easy)
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
随机推荐
- [转载]C# 多线程、控制线程数提高循环输出效率
C#多线程及控制线程数量,对for循环输出效率. 虽然输出不规律,但是效率明显提高. 思路: 如果要删除1000条数据,只使用for循环,则一个接着一个输出.所以,把1000条数据分成seed段,每段 ...
- JavaScript高级---适配器模式
一.设计模式 javascript里面给我们提供了很多种设计模式: 工厂.桥.组合.门面.适配器.装饰者.享元.代理.观察者.命令.责任链 在前面我们实现了工厂模式和桥模式 工厂模式 : 核心:为了生 ...
- [设计模式] 13 责任链模式 Chain of Responsibility
转 http://blog.csdn.net/wuzhekai1985 http://www.jellythink.com/archives/878 向项目经理提交了休假申请,我的项目经理向 ...
- window dos 设置网络
->netsh ->pushd interface ip ->set address "本地连接" static 192.168.1.2 255.255.255. ...
- hdu 1056
水题 ~~ 按题目要求直接判断~. /************************************************************************* > A ...
- c#中HttpWebRequest使用Proxy实现指定IP的域名请求
原文:http://www.cnblogs.com/greenerycn/archive/2010/04/11/httpwebreques_host_modify_By_set_proxy.html ...
- grunt下cssmin的配置参数
每个目标的具体设置,需要参考该模板的文档minify目标的参数具体含义如下: expand:如果设为true,就表示下面文件名的占位符(即*号)都要扩展成具体的文件名. cwd:需要处理的文件(inp ...
- java使用正则表达式——实例
Java代码 import java.util.regex.Matcher; import java.util.regex.Pattern; /** * @author Der * ...
- 【转载】Eclipse自动编译问题
今天调试的时候发现问题:调试的时候竟然在我注释的里面走,当时那个郁闷啊,每次都要clean下才可以,晚上感觉不对劲,上网查了查,原来是bulid automatically这个我把勾去掉了,下面是原文 ...
- NET 使用HtmlAgilityPack抓取网页数据
刚刚学习了XPath路径表达式,主要是对XML文档中的节点进行搜索,通过XPath表达式可以对XML文档中的节点位置进行快速定位和访问,html也是也是一种类似于xml的标记语言,但是语法没有那么严谨 ...