【Leetcode】每日一题：N 叉树的层序遍历

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【Leetcode】每日一题：N 叉树的层序遍历

N 叉树的层序遍历

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）

AC代码

"""
# Definition for a Node.
class Node:def __init__(self, val=None, children=None):self.val = valself.children = children
"""class Solution:def levelOrder(self, root: 'Node') -> List[List[int]]:if root is None:return []result = []queue = [root]while queue != []:temp = []t = copy.deepcopy(queue)for r in t:temp.append(r.val)queue.pop(0)queue += r.childrenresult.append(temp)return result

官方代码

class Solution:def levelOrder(self, root: 'Node') -> List[List[int]]:if not root:return []ans = list()q = deque([root])while q:cnt = len(q)level = list()for _ in range(cnt):cur = q.popleft()level.append(cur.val)for child in cur.children:q.append(child)ans.append(level)return ans# 作者：LeetCode-Solution

1、官方代码用的双端队列，本质其实和笔者的代码是相同的