0993._Cousins_in_Binary_Tree.md

993. Cousins in Binary Tree

难度: Easy

刷题内容

原题连接

• https://leetcode.com/problems/cousins-in-binary-tree/

内容描述

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:

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

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******

遍历，记录parent和depth

class Solution:
    def isCousins(self, root: 'TreeNode', x: 'int', y: 'int') -> 'bool':
        self.parent, self.depth = {}, {}
        
        def dfs(node, par = None):
            if not node:
                return
            self.depth[node.val] = 1 + self.depth[par.val] if par else 0
            self.parent[node.val] = par
            dfs(node.left, node)
            dfs(node.right, node)

        dfs(root)
        return self.depth[x] == self.depth[y] and self.parent[x] != self.parent[y]