【LeetCode】106. Construct Binary Tree from Inorder and Postorder Traversal 解题报告
【LeetCode】106. Construct Binary Tree from Inorder and Postorder Traversal 解题报告(Python)
标签: LeetCode
题目描述:
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
题目大意
根据中序遍历和后序遍历重建二叉树。
解题方法
这个是套路题,我没有完全记住每一步是多少,而是根据树的样子和两个数组进行分析,得出切片的位置。
后序遍历的最后一个元素一定是根节点,在中序遍历中找出此根节点的位置序号。中序遍历序号左边的是左孩子,右边的是右孩子。再根据左孩子和右孩子的长度对后序遍历进行切片即可。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def buildTree(self, inorder, postorder):
"""
:type inorder: List[int]
:type postorder: List[int]
:rtype: TreeNode
"""
if not inorder or not postorder: return None
val = postorder[-1]
root = TreeNode(val)
index = inorder.index(val)
root.left = self.buildTree(inorder[:index], postorder[:index])
root.right = self.buildTree(inorder[index+1:], postorder[index:-1])
return root
日期
2018 年 3 月 12 日
【LeetCode】106. Construct Binary Tree from Inorder and Postorder Traversal 解题报告的更多相关文章
- Java for LeetCode 106 Construct Binary Tree from Inorder and Postorder Traversal
Construct Binary Tree from Inorder and Postorder Traversal Total Accepted: 31041 Total Submissions: ...
- LeetCode: Construct Binary Tree from Inorder and Postorder Traversal 解题报告
Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of ...
- (二叉树 递归) leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that ...
- [LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树
Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that ...
- LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal (用中序和后序树遍历来建立二叉树)
Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that ...
- C#解leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that ...
- LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树 C++
Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that ...
- Leetcode#106 Construct Binary Tree from Inorder and Postorder Traversal
原题地址 二叉树基本操作 [ ]O[ ] [ ][ ]O 代码: TreeNode *restore(vector<i ...
- [leetcode] 106. Construct Binary Tree from Inorder and Postorder Traversal(medium)
原题地址 思路: 和leetcode105题差不多,这道题是给中序和后序,求出二叉树. 解法一: 思路和105题差不多,只是pos是从后往前遍历,生成树顺序也是先右后左. class Solution ...
随机推荐
- 2.MaxSubArray-Leetcode
题目:最大连续子序列和 思路:动态规划 状态转移方程 f[j]=max{f[j-1]+s[j],s[j]}, 其中1<=j<=n target = max{f[j]}, 其中1<=j ...
- 解决windows 10由于签名原因无法安装ADB driver 的问题
ADB Driver Installer (Automatically) In Windows 8 (8.1) or 10 64-bit you are unable to install unsig ...
- CMSIS-RTOS 信号量Semaphores
信号量Semaphores 和信号类似,信号量也是一种同步多个线程的方式,简单来讲,信号量就是装有一些令牌的容器.当一个线程在执行过程中,就可能遇到一个系统调用来获取信号量令牌,如果这个信号量包含多个 ...
- mongoDB整个文件夹拷贝备份还原的坑
现网有一个mongoDB数据库需要搬迁到新服务器,开发那边的要求是先搬迁现在的数据库过去,然后剩下的以后他们用程序同步. 数据库大楷20G左右,现网是主备仲裁的,停掉备点,拷贝了全部文件. 新服务器也 ...
- 学习java的第八天
一.今日收获 1.学习完全学习手册上2.3转义字符与2.4运算符两节 二.今日难题 1.没有什么难理解的问题 三.明日目标 1.哔哩哔哩教学视频 2.Java学习手册
- 巩固javaweb的第三十天
显示用户输入信息 1 .代码 要想输出用户在上一个页面提交的信息,可以使用下面的代码: ${param.userid} ${param.username} ${param.userpass} ${pa ...
- Shell $()、${}、$[]、$(())
目录 Shell中的 $().${}.$[].$(()) $().${} 替换 ${} 变量内容的替换.删除.取代 数组 $[].$(()) 运算符 Shell中的 $().${}.$[].$(()) ...
- 大数据学习day25------spark08-----1. 读取数据库的形式创建DataFrame 2. Parquet格式的数据源 3. Orc格式的数据源 4.spark_sql整合hive 5.在IDEA中编写spark程序(用来操作hive) 6. SQL风格和DSL风格以及RDD的形式计算连续登陆三天的用户
1. 读取数据库的形式创建DataFrame DataFrameFromJDBC object DataFrameFromJDBC { def main(args: Array[String]): U ...
- 【leetcode】633. Sum of Square Numbers(two-sum 变形)
Given a non-negative integer c, decide whether there're two integers a and b such that a2 + b2 = c. ...
- Linux学习 - 脚本安装包
脚本安装包不是独立的软件包类型,常见安装的是源码包