951. 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 if we can make X equal to Y after some number of flip operations.

Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivelent or false otherwise.

 

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.

Example 2:

Input: root1 = [], root2 = []
Output: true

Example 3:

Input: root1 = [], root2 = [1]
Output: false

Example 4:

Input: root1 = [0,null,1], root2 = []
Output: false

Example 5:

Input: root1 = [0,null,1], root2 = [0,1]
Output: true

 

Constraints:

• The number of nodes in each tree is in the range [0, 100].
• Each tree will have unique node values in the range [0, 99].

class Solution {
    public boolean flipEquiv(TreeNode root1, TreeNode root2) {
        if(root1 == null || root2 == null) return root1 == root2;
        if(root1.val != root2.val) return false;
        return (flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right)) ||
                         (flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left)); 
    }
}

牛逼的。。首先跟小编来理解一下flip后相等是怎么回事，好的就是这么回事，你懂了吗？

题目要flip或者不flip后相等，就可以分成两种情况。

首先是终止条件：说起来是flip，实际上还是比较root是不是相等。如果root1或者root2是null了，直接看他俩是否相等。然后因为要相等，所以要判断root1和root2的val。如果不相等直接返回false

然后进入递归，前面我们说了有两种情况。1. 不flip，就返回比较他俩left和right child的情况 2. flip，返回比较root1.left, root2.right，和root1.right, root2.left的情况。

https://leetcode.com/problems/flip-equivalent-binary-trees/discuss/200514/JavaPython-3-DFS-3-liners-and-BFS-with-explanation-time-and-space%3A-O(n).