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  • 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.

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  • 原文地址:https://www.cnblogs.com/captain-dl/p/10176342.html
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