[leetcode tree]101. Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/ \
2 2
\ \
3 3
判断一棵树是否对称
递归:
class Solution(object):
def isSymmetric(self, root):
if not root:
return True
return self.issym(root.left,root.right)
def issym(self,left,right):
if not left and not right:
return True
if not left or not right:
return False
if left.val == right.val:
return self.issym(left.left,right.right) and self.issym(left.right,right.left)
return False
非递归:
递归本质上就是栈,所以。。。
class Solution(object):
def isSymmetric(self, root):
if not root:
return True
stack=[[root.left,root.right]]
while stack:
left,right = stack.pop()
if not left and not right:
continue
if not left or not right:
return False
if left.val == right.val:
stack.append([left.right,right.left])
stack.append([left.left,right.right])
else:
return False
return True
[leetcode tree]101. Symmetric Tree的更多相关文章
- Leetcode之101. Symmetric Tree Easy
Leetcode 101. Symmetric Tree Easy Given a binary tree, check whether it is a mirror of itself (ie, s ...
- Leetcode 笔记 101 - Symmetric Tree
题目链接:Symmetric Tree | LeetCode OJ Given a binary tree, check whether it is a mirror of itself (ie, s ...
- 【LeetCode】101 - Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
- LeetCode OJ 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 (2 solutions)
Symmetric Tree Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its ...
- (Tree) 101. Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For e ...
- 【一天一道LeetCode】#101. Symmetric Tree
一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given a ...
- 【LeetCode】101. Symmetric Tree 对称二叉树(Java & Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS BFS 日期 [LeetCode] 题目地址 ...
- [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 ...
随机推荐
- 基于受限玻尔兹曼机(RBM)的协同过滤
受限玻尔兹曼机是一种生成式随机神经网络(generative stochastic neural network), 详细介绍可见我的博文<受限玻尔兹曼机(RBM)简介>, 本文主要介绍R ...
- BZOJ4816 数字表格
4816: [Sdoi2017]数字表格 Time Limit: 50 Sec Memory Limit: 128 MB Description Doris刚刚学习了fibonacci数列.用f[i ...
- Google Congestion Control介绍
随着网络带宽的日益增加和便携式设备,如智能手机或平板电脑处理能力的增强,基于互联网的实时通信已经成为热点. 虽然视频会议已商用了多年,特别是SKYPE这样的视频应用在互联网上已有10年时间,但针对实时 ...
- Linux的基础优化-2
1.启动网卡 ifup eth0 2.SSH链接 ifconfig 查看IP后SSH终端连接3.更新源 最小化安装是没有wget工具的,必须先安装再修改源 yum install wget 备份原系统 ...
- c# 生成随机N位数字串(每位都不重复)
/// <summary> /// 生成随机数字窜 /// </summary> /// <param name="Digit">位数</ ...
- [Openwrt 扩展下篇] Openwrt搭建私有云Owncloud 9
网上很多资料讲用Linux打造owncloud构建私有云 ,花了些时间研究了下,我将之前的需求打造成了Openwrt下的Owncloud 9.其实网上还有Seafile.大家对比来看下知乎的评论,其实 ...
- const与指针
C++中const与指针 1.常指针: ; int * const pInt = &x; 其中PInt是常指针,pInt的值无法改变,但其指向的内容可以改变. 2.指向常量的指针 有两种写法: ...
- sql_injection之post注入
1.代码篇 </html> <center> <form action="#" method="post"> 姓名:< ...
- java四舍五入BigDecimal和js保留小数点两位
java四舍五入BigDecimal保留两位小数的实现方法: // 四舍五入保留两位小数System.out.println("四舍五入取整:(3.856)=" + ne ...
- 搜索入门之dfs--经典的迷宫问题解析
今天来谈一下dfs的入门,以前看到的dfs入门,那真的是入门吗,都是把dfs的实现步骤往那一贴,看完是知道dfs的步骤了,但是对于代码实现还是没有概念.今天准备写点自己的心得,真的是字面意思--入门. ...