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

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].

我們可以為二叉樹 T 定義一個翻轉操作，如下所示：選擇任意節點，然後交換它的左子樹和右子樹。

只要經過一定次數的翻轉操作後，能使 X 等於 Y，我們就稱二叉樹 X 翻轉等價於二叉樹 Y。

編寫一個判斷兩個二叉樹是否是翻轉等價的函式。這些樹由根節點 root1和 root2 給出。

示例：

輸入：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]
輸出：true
解釋：We flipped at nodes with values 1, 3, and 5.

提示：

1. 每棵樹最多有 100 個節點。
2. 每棵樹中的每個值都是唯一的、在 [0, 99] 範圍內的整數。

16ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func flipEquiv(_ root1: TreeNode?, _ root2: TreeNode?) -> Bool {
16         if root1 == nil {return root2 == nil}
17         if root2 == nil {return root1 == nil}
18         if root1!.val != root2!.val {return false}
19         if flipEquiv(root1!.left, root2!.left) && flipEquiv(root1!.right, root2!.right)
20         {
21             return true
22         }
23         if flipEquiv(root1!.left, root2!.right) && flipEquiv(root1!.right, root2!.left)
24         {
25             return true
26         }
27         return false
28     }
29 }