LeetCode 993 Cousins in Binary Tree 解题报告

题目要求

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.

题目分析及思路

定义二叉树的深度：根结点的深度为0，深度为k的结点的孩子结点的深度为k+1。如果二叉树里的两个结点有相同的深度和不同的父结点，则称这两个结点为cousins。给定二叉树的根结点（每个结点的值唯一）以及两个不同结点的值，判断这两个结点是否是cousins。可以使用堆来存储结点，用字典parent和depth来存储各个结点的深度和父结点的值。遍历树的每一层，对该层结点的深度和父结点进行记录。

python代码

# Definition for a binary tree node.

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

def isCousins(self, root: TreeNode, x: int, y: int) -> bool:

parent = {}

depth = {}

q = collections.deque()

q.append(root)

k = -1

while q:

k += 1

size = len(q)

for _ in range(size):

node = q.popleft()

if not node:

continue

depth[node.val] = k

q.append(node.left)

if node.left:

parent[node.left.val] = node.val

q.append(node.right)

if node.right:

parent[node.right.val] = node.val

return depth[x] == depth[y] and parent[x] != parent[y]