
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:
The depth of the tree is at most1000. 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:
1 """ 2 # DeFinition for a Node. 3 class Node: 4 def __init__(self,val,children): 5 self.val = val 6 self.children = children 7 """ 8 class Solution: 9 def levelOrder(self,root: ‘Node‘) -> List[List[int]]:10 if not root:11 return []12 13 result,queue = [],[(root,1)]14 while queue:15 Node,level = queue.pop()16 if level - len(result) >= 1:17 result.append([])18 result[level - 1].append(Node.val)19 20 for child in Node.children:21 queue.insert(0,(child,level+ 1))22 return result总结
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