【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 = 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