题目如下:

Given a binary tree with the following rules:

  1. root.val == 0
  2. If treeNode.val == x and treeNode.left != null, then treeNode.left.val == 2 * x + 1
  3. If treeNode.val == x and treeNode.right != null, then treeNode.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 the target 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 True

Example 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 False

Example 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 True

Constraints:

  • 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

  1. 【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 ...

  2. 【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:/ ...

  3. 【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 ...

  4. 【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 ...

  5. 【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 ...

  6. 【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 ...

  7. 【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 ...

  8. 【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 ...

  9. 【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

  1. PhpMyAdmin提示未加密

    新版本的PhpMyAdmin 增强了安全性,需要在配置文件设置一个短语密码.否则进入之后会有“配置文件现在需要一个短语密码.”的红色警叹提示. 解决方法: .将 phpMyAdmin/librarie ...

  2. Emgu 学习之HelloWorld

    安装和配置 系统Win10,VS2013,下载Emgu安装包libemgucv-windesktop-3.4.3.3016 安装到了E:\OpenCV\emgucv-windesktop 3.4.3. ...

  3. python常用关键字

    转载: python关键字以及含义,用法 查看所有python关键字: import keyword print(keyword.kwlist) ['False', 'None', 'True', ' ...

  4. hue的优化

    参考: 官网: https://docs.cloudera.com/documentation/enterprise/6/6.2/topics/hue_ref_arch.html 1/ 和开发沟通是否 ...

  5. CDH的mysql主从准备

    参考: https://www.cnblogs.com/yinzhengjie/p/10371899.html https://www.sysit.cn/blog/post/sysit/CDH6.2. ...

  6. sql server2012中使用convert来取得datetime数据类型样式(全)

    一.日期数据格式的处理,两个示例: CONVERT(varchar(16), 时间一, 20) 结果:2007-02-01 08:02/*时间一般为getdate()函数或数据表里的字段*/ CONV ...

  7. 算法 - k-means算法

    一.聚类思想 所谓聚类算法是指将一堆没有标签的数据自动划分成几类的方法,属于无监督学习方法,这个方法要保证同一类的数据有相似的特征,如下图所示:     根据样本之间的距离或者说是相似性(亲疏性),把 ...

  8. IDEA项目目录里下找不到src,但是src确实存在的的解决方案

    写代码的时候可能出现写着写着src就找不到了,我个人认为是触发了热键导致src被隐藏了,下面就是设置src可见和不可见的操作 这个其实是被隐藏了,打开就好,位置如下:

  9. CentOS 7 配置 kcptun 实现网站加速

    目的:shadowsocks+kcptun 实现vpn加速(shadowsocks,kcptun在同一台VPS上) 一.shadowsocks安装(参考  https://www.cnblogs.co ...

  10. weex animation模块 使用指南

    本节学习目标 掌握内置组件animation的使用 我们在开发应用的时候,常常需要增加一些动画效果,来提高用户体验,经常用到的一些动画效果如下 平移 旋转 缩放 背景颜色改变 组件透明图 weex 提 ...