LeetCode | 0429. N 叉树的层序遍历【Python】

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

posted @ 2021-01-11 21:45  Wonz  阅读(149)  评论(0编辑  收藏  举报