LeetCode OJ--Construct Binary Tree from Inorder and Postorder Traversal *
http://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
知道二叉树的中序遍历和后序遍历序列,求二叉树。
使用递归
#include <iostream>
#include <vector>
using namespace std; struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
}; class Solution{
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder)
{
return buildTree(begin(inorder),end(inorder),begin(postorder),end(postorder));
}
template<typename BidiIt>
TreeNode* buildTree(BidiIt in_first, BidiIt in_last,BidiIt post_first,BidiIt post_last)
{
if(in_first ==in_last)
return nullptr;
if(post_first == post_last)
return nullptr;
const auto val = *prev(post_last);
TreeNode* root = new TreeNode(val); auto in_root_pos = find(in_first,in_last,val);
auto left_size = distance(in_first,in_root_pos);
auto post_left_last = next(post_first,left_size); root->left = buildTree(in_first,in_root_pos,post_first,post_left_last);
root->right = buildTree(next(in_root_pos),in_last,post_left_last,prev(post_last));
return root;
}
}; int main()
{
Solution myS;
int arr1[] = {,};//,5,1,6,3,7 };
int arr2[] = {,};//,2,6,7,3,1 };
vector<int> inorder(arr1,arr1+) ;
vector<int> postorder(arr2 ,arr2+);
TreeNode *myNode; myNode = myS.buildTree(inorder,postorder); cout<<"hi"<<endl;
return ;
}
LeetCode OJ--Construct Binary Tree from Inorder and Postorder Traversal *的更多相关文章
- [Leetcode Week14]Construct Binary Tree from Inorder and Postorder Traversal
Construct Binary Tree from Inorder and Postorder Traversal 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/pr ...
- 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 -day23 Construct Binary Tree from Inorder and Postorder Traversal & Construct Binary Tree f
1. Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder travers ...
- (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】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[105] Construct Binary Tree from Inorder and Postorder Traversal
代码实现:给定一个中序遍历和后序遍历怎么构造出这颗树!(假定树中没有重复的数字) 因为没有规定是左小右大的树,所以我们随意画一颗数,来进行判断应该是满足题意的. 3 / \ 2 4 /\ / \1 6 ...
随机推荐
- Python基础——判断和循环
判断 缩进代替大括号. 冒号(:)后换号缩进. if test=100 if test>50: print('OK') print('test') if-elif-else test=50 if ...
- STM32F407VET6之IAR之ewarm7.80.4工程建立(基于官方固件库1.6版本) 的工程文件目录
最后整理结构如下所示,├─cmsis│ startup_stm32f401xx.s│ startup_stm32f40xx.s│ startup_stm32f40_41xxx.s│ startup_s ...
- ubuntu12.04ppa安装emacs24
ppa地址:https://launchpad.net/~cassou/+archive/emacs 因为debian版本的emacs-snapshot维护者停止更新,所有ubuntu上的也停止了. ...
- linux学习-用户的特殊 shell 与 PAM 模块
特殊的 shell, /sbin/nologin 『无法登入』指的是:『这个使用者无法使用 bash 或其他 shell 来登入系统』而已, 并不是说这个账号就无法使用其他的系统资源! 让某个具有 / ...
- 重新造轮子之静态链接1(Static linking)
最近学习计算机病毒学的过程中,又讲到了静态链接的问题,联想到了之前保健哥在信息安全的课堂上向我们展示了一个没有main()函数的C程序到底应该如何编写.个人觉得这个小实验对于加深静态链接的过程的理解也 ...
- HDU 5527 Too Rich 贪心
题意: 有\(10\)种面值为\(1, 5, 10, 20, 50, 100, 200, 500, 1000, 2000\)的纸币,现在你要选最多的数量凑成\(p\)块钱. 分析: 同样分析问题的反面 ...
- 网页静态化解决方案Freemarker
序言: 沉淀了三个月,逐步将自己最近两年在公司中用到的技术和知识点,重新整理归纳了下,对比以前可以发现,现在技术更新越来越快,也越来越成熟,在互联网企业,用到的技术也更先进,更领先,比如微服务.分布式 ...
- adb -a server nodaemon,设备一直显示 offline,而 adb devices 一直显示 device【已解决】
1. adb -a server nodaemon 一直显示 offline 2. adb devices 一直显示 device 谷歌 和 度娘了一圈,未寻得解决办法 # 解决方法 问题已解决,使用 ...
- 转:模式窗口showModalDialog的用法总结
模式窗口showModalDialog的用法总结 最近几天一直在处理模式窗口的问题,索性写了这篇总结,以供参考: 1.打开窗口:var handle = window.showModalDialo ...
- Python 单例模式(3种方式)
方式一: # 单例模式: # 实现目的:实例化多次,得到的实例是同一个,就是同一个对象,同一个名称空间(更加节省空间) ####################################方式一: ...