[Swift Weekly Contest 113]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 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,7,8],root2 = [1,8,7] Output: true Explanation: We flipped at nodes with values 1,and 5. 
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Note:
 

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


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,root2 = [1,7]
输出：true
解释：We flipped at nodes with values 1,and 5.
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提示：
 

每棵树最多有 100 个节点。
 
 
每棵树中的每个值都是唯一的、在 [0,99] 范围内的整数。
 


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 }@H_404_30@