【leetcode】1261. Find Elements in a Contaminated Binary Tree
题目如下:
Given a binary tree with the following rules:
root.val == 0
- If
treeNode.val == x
andtreeNode.left != null
, thentreeNode.left.val == 2 * x + 1
- If
treeNode.val == x
andtreeNode.right != null
, thentreeNode.right.val == 2 * x + 2
Now the binary tree is contaminated, which means all
treeNode.val
have been changed to-1
.You need to first recover the binary tree and then implement the
FindElements
class:
FindElements(TreeNode* root)
Initializes the object with a contamined binary tree, you need to recover it first.bool find(int target)
Return if thetarget
value exists in the recovered binary tree.Example 1:
Input
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
Output
[null,false,true]
Explanation
FindElements findElements = new FindElements([-1,null,-1]);
findElements.find(1); // return False
findElements.find(2); // return TrueExample 2:
Input
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
Output
[null,true,true,false]
Explanation
FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
findElements.find(1); // return True
findElements.find(3); // return True
findElements.find(5); // return FalseExample 3:
Input
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
Output
[null,true,false,false,true]
Explanation
FindElements findElements = new FindElements([-1,null,-1,-1,null,-1]);
findElements.find(2); // return True
findElements.find(3); // return False
findElements.find(4); // return False
findElements.find(5); // return TrueConstraints:
TreeNode.val == -1
- The height of the binary tree is less than or equal to
20
- The total number of nodes is between
[1, 10^4]
- Total calls of
find()
is between[1, 10^4]
0 <= target <= 10^6
解题思路:题目很简单,先把树恢复,然后判断值是否存在。
代码如下:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None class FindElements(object):
dic = {}
root = None
def __init__(self, root):
"""
:type root: TreeNode
"""
self.dic = {}
self.root = root
def recursive(node,node_val):
node.val = node_val
self.dic[node.val] = 1
if node.left != None:
recursive(node.left,node.val*2+1)
if node.right != None:
recursive(node.right,node.val*2+2)
recursive(self.root,0) def find(self, target):
"""
:type target: int
:rtype: bool
"""
return target in self.dic # Your FindElements object will be instantiated and called as such:
# obj = FindElements(root)
# param_1 = obj.find(target)
【leetcode】1261. Find Elements in a Contaminated Binary Tree的更多相关文章
- 【LeetCode】863. All Nodes Distance K in Binary Tree 解题报告(Python)
[LeetCode]863. All Nodes Distance K in Binary Tree 解题报告(Python) 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http ...
- 【LeetCode】331. Verify Preorder Serialization of a Binary Tree 解题报告(Python)
[LeetCode]331. Verify Preorder Serialization of a Binary Tree 解题报告(Python) 标签: LeetCode 题目地址:https:/ ...
- 【LeetCode】236. Lowest Common Ancestor of a Binary Tree 解题报告(Python)
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 日期 题目地址:https://leetcode.c ...
- 【LeetCode】987. Vertical Order Traversal of a Binary Tree 解题报告(C++ & Python)
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 DFS BFS 日期 题目地址:https://le ...
- 【LeetCode】1161. Maximum Level Sum of a Binary Tree 解题报告 (C++)
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 日期 题目地址:https://leetcod ...
- 【LeetCode】671. Second Minimum Node In a Binary Tree 解题报告(Python)
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 找出所有值再求次小值 遍历时求次小值 日期 题目地址 ...
- 【LeetCode】236. Lowest Common Ancestor of a Binary Tree
Lowest Common Ancestor of a Binary Tree Given a binary tree, find the lowest common ancestor (LCA) o ...
- 【leetcode】1161. Maximum Level Sum of a Binary Tree
题目如下: Given the root of a binary tree, the level of its root is 1, the level of its children is 2, a ...
- 【leetcode】331. Verify Preorder Serialization of a Binary Tree
题目如下: One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null ...
随机推荐
- 【OpenGL】初识OpenGL4.0
目录(?)[-] 什么是GLSL GLEW 安装GLEW 使用GLEW 其他库 使用GLM库进行数学运算 安装GLM 使用GLM 使用GLM作为OpenGL的输入 使用GLFW进行窗口管理 这篇文章主 ...
- 【GO】一个容易踩坑的内外变量屏蔽问题
package main import ( "errors" "fmt" ) func et()(string,error){ return "&qu ...
- 使用feign上传图片
1.添加依赖,支持SpringEncoder <dependency> <groupId>io.github.openfeign.form</groupId> &l ...
- [转帖]2018年的新闻: 国内首家!腾讯主导Apache Hadoop新版本发布
国内首家!腾讯主导Apache Hadoop新版本发布 https://blog.csdn.net/weixin_34194317/article/details/88811258 腾讯也挖了很多 ...
- Git和Github的使用
Git和Github的基本操作 一.了解Git和Github 1.什么是GIT? Git是一个免费.开源的版本控制软件 2.什么是版本控制系统? 版本控制是一种记录一个或若干个文件内容变化,以便将来查 ...
- Spring(十三)-- Spring 事务
Spring 事务 1. 回忆之前事务知识点 一:事务的概念 将一系列的数据操作捆绑在一起,成为一个整体进行统一管理! 一条或者多条sql语句的集合! 二:事务的ACID特性 原子性(Atomic ...
- 【转】.NET Core + Ocelot + IdentityServer4 + Consul 基础架构实现
作者:Zhang_Xiang 原文地址:.NET Core + Ocelot + IdentityServer4 + Consul 基础架构实现 先决条件 关于 Ocelot 针对使用 .NET 开发 ...
- 解决Eclipse发布到Tomcat丢失依赖jar包的问题
解决Eclipse发布到Tomcat丢失依赖jar包的问题 如果jar文件是以外部依赖的形式导入的.Eclipse将web项目发布到Tomcat时,是不会自动发布这些依赖的. 可以通过Eclipse在 ...
- Luogu P3810 【模板】三维偏序(陌上花开)(CDQ分治)
题目 以三维偏序为例来讲一下CDQ分治. CDQ的本质就是把一个序列分成两段,计算左边对右边的贡献,然后分治. 不过一般都是先分治到底再从下往上算,这样可以先归并再算. 比如这道题,我们先按第一维排序 ...
- 如何實現输入字符串This is an Apple on eBay 输出 Siht si na Elppa no yAbe
<?php $str = "This is an Apple on eBay"; //定义字符串 $len = strlen($str); //字符串长度 $sup = [] ...