[Swift Weekly Contest 124]LeetCode993. 二叉树的堂兄弟节点 | Cousins in Binary Tree

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

Note:

1. The number of nodes in the tree will be between 2 and 100.
2. Each node has a unique integer value from 1 to 100.

如果二叉树的两个节点深度相同，但父节点不同，则它们是一对堂兄弟节点。

我们给出了具有唯一值的二叉树的根节点 root，以及树中两个不同节点的值 x 和 y。

只有与值 x 和 y 对应的节点是堂兄弟节点时，才返回 true。否则，返回 false。

示例 1：

输入：root = [1,2,3,4], x = 4, y = 3
输出：false

示例 2：

输入：root = [1,2,3,null,4,null,5], x = 5, y = 4
输出：true

示例 3：

输入：root = [1,2,3,null,4], x = 2, y = 3
输出：false 

提示：

1. 二叉树的节点数介于 2 到 100 之间。
2. 每个节点的值都是唯一的、范围为 1 到 100 的整数。

Runtime: 12 ms

Memory Usage: 18.4 MB

 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     var ddd:Int = 0
16     func isCousins(_ root: TreeNode?, _ x: Int, _ y: Int) -> Bool {
17         var px:TreeNode? = dfs(root, nil, x, 0, 1)
18         var py:TreeNode? = dfs(root, nil, y, 0, -1)
19         return x != y && px != nil && py != nil && px!.val != py!.val && ddd == 0
20     }
21
22     func dfs(_ cur:TreeNode?,_ par:TreeNode?,_ x:Int,_ dep:Int,_ mul:Int) -> TreeNode?
23     {
24         if cur == nil {return nil}
25         if cur!.val == x
26         {
27             ddd += dep * mul
28             return par
29         }
30         var res:TreeNode? = dfs(cur!.left, cur, x, dep + 1, mul)
31         if res != nil {return res}
32         res = dfs(cur!.right, cur, x, dep+1, mul)
33         return res
34     }
35 }

原文地址：https://www.cnblogs.com/strengthen/p/10390643.html