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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