For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

Write a function that determines whether two binary trees are flip equivalent.  The trees are given by root nodes root1 and root2.

Example 1:

Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.

Note:

  1. Each tree will have at most 100 nodes.
  2. Each value in each tree will be a unique integer in the range [0, 99].

A、B两颗二叉树相等当且仅当rootA->data == rootB->data,且A、B的左右子树相等或者左右互换相等

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool flipEquiv(TreeNode* root1, TreeNode* root2) {
if (!root1 && !root2)
return ;
if ((!root1&&root2) || (root1 && !root2))
return ;
if (root1&&root2)
{
if (root1->val == root2->val)
{
if (flipEquiv(root1->left, root2->left))
return flipEquiv(root1->right, root2->right);
else if (flipEquiv(root1->left, root2->right))
return flipEquiv(root1->right, root2->left);
}
}
return ;
}
};

113th LeetCode Weekly Contest Flip Equivalent Binary Trees的更多相关文章

  1. 【LeetCode】951. Flip Equivalent Binary Trees 解题报告(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 递归 日期 题目地址:https://leetcod ...

  2. 【leetcode】951. Flip Equivalent Binary Trees

    题目如下: For a binary tree T, we can define a flip operation as follows: choose any node, and swap the ...

  3. #Leetcode# 951. Flip Equivalent Binary Trees

    https://leetcode.com/problems/flip-equivalent-binary-trees/ For a binary tree T, we can define a fli ...

  4. [Swift]LeetCode951. 翻转等价二叉树 | Flip Equivalent Binary Trees

    For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left a ...

  5. 113th LeetCode Weekly Contest Reveal Cards In Increasing Order

    In a deck of cards, every card has a unique integer.  You can order the deck in any order you want. ...

  6. 113th LeetCode Weekly Contest Largest Time for Given Digits

    Given an array of 4 digits, return the largest 24 hour time that can be made. The smallest 24 hour t ...

  7. Leetcode951. Flip Equivalent Binary Trees翻转等价二叉树

    我们可以为二叉树 T 定义一个翻转操作,如下所示:选择任意节点,然后交换它的左子树和右子树. 只要经过一定次数的翻转操作后,能使 X 等于 Y,我们就称二叉树 X 翻转等价于二叉树 Y. 编写一个判断 ...

  8. leetcode_951. Flip Equivalent Binary Trees_二叉树遍历

    https://leetcode.com/problems/flip-equivalent-binary-trees/ 判断两棵二叉树是否等价:若两棵二叉树可以通过任意次的交换任意节点的左右子树变为相 ...

  9. leetcode weekly contest 43

    leetcode weekly contest 43 leetcode649. Dota2 Senate leetcode649.Dota2 Senate 思路: 模拟规则round by round ...

随机推荐

  1. linux Shell中常用的条件判断

    linux Shell中常用的条件判断 -b file            若文件存在且是一个块特殊文件,则为真 -c file            若文件存在且是一个字符特殊文件,则为真 -d ...

  2. linux下方便的录屏命令

    linux下方便的录屏命令   ffmpeg -f x11grab -s 1024*768 -r 20 -i :0.0 -sameq ~/recode.mpg -r后是刷新屏率,   推出直接Ctrl ...

  3. Linux扩展根目录

    一.简介 使用linux系统的过程中,有时发现系统根目录(/)的空间不足,导致系统运行很慢,针对该现象,本文详细介绍根目录(/)的空间扩展方法.   二.操作步骤 1)查看根目录大小 df 2)查找系 ...

  4. p2148 [SDOI2009]E&D

    传送门 分析 https://www.luogu.org/blog/flashblog/solution-p2148 代码 #include<bits/stdc++.h> using na ...

  5. ShopNc登录

  6. Introduction to Partial View

    By Jignesh Trivedi on May 14, 2015 http://www.c-sharpcorner.com/UploadFile/ff2f08/partial-view-in-mv ...

  7. Java SimpleDateFormat工具类

    package AnimalDemo; import java.text.ParseException; import java.text.SimpleDateFormat; import java. ...

  8. 删除一个数的K位使原数变得最小

    原创 给定一个n位正整数a, 去掉其中k个数字后按原左右次序将组成一个新的正整数.对给定的a, k寻找一种方案,使得剩下的数字组成的新数最小. 提示:应用贪心算法设计求解 操作对象为n位正整数,有可能 ...

  9. android开关控件Switch和ToggleButton

    序:今天项目中用到了开关按钮控件,查阅了一些资料特地写了这篇博客记录下. 1.Switch <Switch android:id="@+id/bt" android:layo ...

  10. TCP连接状态-如何判断一个TCP连接是否可用

    在使用一个长连接的TCP时,如果TCP服务器端接收到TCP的客户端连接过来后,接着服务器端的TCP节点需要对这个客户端进行数据收发,收发时需要判断这个SOCKET是否可用用,判断方法有多种: 1.li ...