作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 BFS DFS 日期 题目地址:https://leetcode.com/problems/binary-tree-level-order-traversal/ 题目描述 Given a binary tree, return the level order traversal of its nodes' values. (ie, from left
话不多说,直接上代码 class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None class Solution: #前序 def preorder(self,root,ans=[]): if root!=None: ans.append(root.val) if root.left: self.preorder(root.left,ans) if root.right: self.p
__author__ = "WSX" class Node: def __init__(self, val = None, left = None, right = None): self.val = val self.left = left self.right = right class tree: def __init__(self, L): self.val = L def bulid(self,root, i): #将列表转化建立二叉树 if i < len(self.
bytes.bytearray #思考下面例子: a = 1 b = a print(a == b)#True print(a is b)#True print(id(a) is id(b))#False print(id(a) == id(b))#True 1. id():获取的是对象在内存中的地址 2. is :比对2个变量的对象引用(对象在内存中的地址,即id() 获得的值)是否相同.如果相同则返回True,否则返回False.换句话说,就是比对2个变量的对象引用是否指向同一个对象. 3.