zoukankan      html  css  js  c++  java
  • LeetCode429. N-ary Tree Level Order Traversal

    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.

  • 相关阅读:
    Working with WordprocessingML documents (Open XML SDK)
    How to Choose the Best Way to Pass Multiple Models in ASP.NET MVC
    Azure:Manage anonymous read access to containers and blobs
    Convert HTML to PDF with New Plugin
    location.replace() keeps the history under control
    On the nightmare that is JSON Dates. Plus, JSON.NET and ASP.NET Web API
    HTTP Modules versus ASP.NET MVC Action Filters
    解读ASP.NET 5 & MVC6系列(6):Middleware详解
    Content Negotiation in ASP.NET Web API
    Action Results in Web API 2
  • 原文地址:https://www.cnblogs.com/captain-dl/p/10176342.html
Copyright © 2011-2022 走看看