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 = val
self.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.children
result.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
