Description
Given an n-ary tree, return the level order traversal of its nodes' values.
Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
Example 1:
Input: root = [1,null,3,2,4,null,5,6] Output: [[1],[3,2,4],[5,6]] Example 2:
Example 2:
Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14] Output: [[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
Constraints:
The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 10^4]
描述
给定一个 N 叉树，返回其节点值的层序遍历。 (即从左到右，逐层遍历)。
例如，给定一个 3叉树 :
返回其层序遍历:
[ [1], [3,2,4], [5,6] ]
说明:
树的深度不会超过 1000。
树的节点总数不会超过 5000。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
思路
- 层次遍历，使用队列。
- 将当前层的所有节点押入队中，记下当前队列中元素的个数。
- for 循环取出当前层的元素，每取一个元素，将其的 children 元素押入队列尾部。
# -*- coding: utf-8 -*- # @Author: 何睿 # @Create Date: 2019-12-28 21:23:06 # @Last Modified by: 何睿 # @Last Modified time: 2019-12-28 21:34:42 from typing import List from collections import deque class Solution: def levelOrder(self, root: 'Node') -> List[List[int]]: if not root: return [] result = [] queue = deque([root]) while queue: tmp = [] count = len(queue) for _ in range(count): node = queue.popleft() tmp.append(node.val) if node.children: queue.extend(node.children) result.append(tmp) return result
源代码文件在 这里 。
