Problem
LeetCode
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:
img
Input: root = [1,null,3,2,4,null,5,6]
Output: [[1],[3,2,4],[5,6]]
Example 2:
img
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, 104]
问题
力扣
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）。
树的序列化输入是用层序遍历，每组子节点都由 null 值分隔（参见示例）。
示例 1：
img
输入：root = [1,null,3,2,4,null,5,6]
输出：[[1],[3,2,4],[5,6]]
示例 2：
img
输入：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]
输出：[[1],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]
提示：
- 树的高度不会超过 1000
- 树的节点总数在 [0, 10^4] 之间
思路
BFS
二叉树层次遍历用 BFS 的时候每次都是把左右子树加入队列，N叉树每次都是把所有子树加入队列。
Python3 代码
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def levelOrder(self, root: 'Node') -> List[List[int]]:
import collections
res = []
if not root:
return res
q = collections.deque()
q.append(root)
# BFS
while q:
tmp = []
for _ in range(len(q)):
node = q.popleft()
tmp.append(node.val)
# 使用extend在列表末尾一次追加多个值
q.extend(node.children)
res.append(tmp)
return res
GitHub 链接
Python