LeetCode429. N-ary Tree Level Order Traversal
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题目来源:429. N-ary Tree Level Order Traversal
https://leetcode.com/problems/n-ary-tree-level-order-traversal/
自我感觉难度/真实难度:hard/easy
队列操作不熟悉
题意:
层序遍历树
分析:
自己的代码:
""" # Definition for a Node. class Node(object): def __init__(self, val, children): self.val = val self.children = children """ class Solution(object): def levelOrder(self, root): """ :type root: Node :rtype: List[List[int]] """ if not root: return [] res=[] temp=[] for i,j in enumerate(root): res.append([i]) temp.push(j) levelOrder(temp) return res
代码效率/结果:
优秀代码:
""" # Definition for a Node. class Node(object): def __init__(self, val, children): self.val = val self.children = children """ class Solution(object): def levelOrder(self, root): """ :type root: Node :rtype: List[List[int]] """ res = [] que = collections.deque() que.append(root) while que: level = [] size = len(que) for _ in range(size): node = que.popleft() if not node: continue level.append(node.val) for child in node.children: que.append(child) if level: res.append(level) return res
代码效率/结果:
自己优化后的代码:
""" # Definition for a Node. class Node(object): def __init__(self, val, children): self.val = val self.children = children """ class Solution(object): def levelOrder(self, root): """ :type root: Node :rtype: List[List[int]] """ if not root: return [] res=[] que=collections.deque() que.append(root) while que: temp=[] #每次要使用的临时变量 size=len(que) #队列的循环通过size来实现 for _ in range(size): node=que.pop() if not node: continue temp.append(node.val) for child in node.children: que.append(child) if temp: res.append(temp) return res
反思改进策略:
1.对队列的操作不熟悉
que=collections.deque() #队列的构造
while que:
temp=[] #每次要使用的临时变量
size=len(que) #队列的循环通过size来实现
for _ in range(size):
node=que.pop() #通过弹出前面的元素来实现, 队列长这个样子:que(),里面放一个长的list
2.
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