LeetCode 951. Flip Equivalent Binary Trees

原题链接在这里：https://leetcode.com/problems/flip-equivalent-binary-trees/

题目：

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].

题解：

Compare root1 and root2, see if it is equal first.

If yes, then

Case 1: root1.left equivalent to root2.left && root1.right equivalent to root2.right

Case 2: root1.left equivalent to root2.right && root1.right equivalent to root2.left.

Either case is true, then return true.

Time Complexity: O(n^2). T(n) = 4*T(n/2) + 1. Master Theorem, T(n) = O(n^2).

Space: O(h).

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public boolean flipEquiv(TreeNode root1, TreeNode root2) {
12         if(root1 == null && root2 == null){
13             return true;
14         }
15
16         if(root1 == null || root2 == null){
17             return false;
18         }
19
20         if(root1.val != root2.val){
21             return false;
22         }
23
24         return (flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right))
25             || (flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left));
26     }
27 }

原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/11072771.html