问题描述:

Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example, given a 3-ary tree:

We should return its level order traversal:

[
[1],
[3,2,4],
[5,6]
]

Note:

  1. The depth of the tree is at most 1000.
  2. The total number of nodes is at most 5000.

思路:此题是在leetcode上见到的第一个BFS的题目,因而记录。同时注意该句代码应用方式 Node,level = queue.pop()

代码:

 """
# Definition for a Node.
class Node:
def __init__(self, val, children):
self.val = val
self.children = children
"""
class Solution:
def levelOrder(self, root: 'Node') -> List[List[int]]:
if not root:
return [] result, queue = [],[(root,1)]
while queue:
Node,level = queue.pop()
if level - len(result) >= 1:
result.append([])
result[level - 1].append(Node.val) for child in Node.children:
queue.insert(0,(child,level+ 1))
return result

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