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. 编写一个判断两个二叉树是否是翻转等价的函数.这些树由根节点 root1 和 root2 给出. 示例: 输入: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] 输出:true…

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 opera…

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://leetcode.com/problems/flip-equivalent-binary-trees/description/ 题目描述 For a binary tree T, we can define a flip operation as follows: choose any node, and swa…

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 opera…

https://leetcode.com/problems/flip-equivalent-binary-trees/ 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…

题目如下: 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…

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同,则称两棵二叉树等价. 思路:遍历二叉树,判断所有的子树是否等价. struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} };…

问题: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. 编写一个判断两个二叉树是否是翻转等价的函数.这些树由根节点 root1 和 root2 给出. 示例: 输入: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,…

Invert a binary tree 翻转一棵二叉树 假设有如下一棵二叉树: 4  / \   2    7  / \   / \ 1  3 6  9翻转后: 4     /    \    7     2   / \    / \  9  6  3  1 这里采用递归的方法来处理.遍历结点,将每个结点的两个子结点交换位置即可. 从左子树开始,层层深入,由底向上处理结点的左右子结点:然后再处理右子树 全部代码如下: public class InvertBinaryTree { public…

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then…