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

  1. [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 ...

  2. 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: ...

  3. leetcode -day23 Construct Binary Tree from Inorder and Postorder Traversal &amp; Construct Binary Tree f

    1.  Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder travers ...

  4. (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 ...

  5. [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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 【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 ...

  10. 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

  1. poj3525 Most Distant Point from the Sea

    题目描述: vjudge POJ 题解: 二分答案+半平面交. 半径范围在0到5000之间二分,每次取$mid$然后平移所有直线,判断半平面交面积是否为零. 我的eps值取的是$10^{-12}$,3 ...

  2. Voyager下的Dashboard Widgets

    widgets设置,voyager.php下找到'widgets': 'widgets' => [ 'TCG\\Voyager\\Widgets\\UserDimmer', 'TCG\\Voya ...

  3. FastJsonUtils工具类

    fastjson是由alibaba开源的一套json处理器.与其他json处理器(如Gson,Jackson等)和其他的Java对象序列化反序列化方式相比,有比较明显的性能优势. 版权声明:本文为博主 ...

  4. urllib、requests库整理

  5. Eclipse设置C++自动补全变量名快捷键

    用快捷键:Alt+/ 要是还是有些场合不能提示,按照下列步骤 Window-Preferences-c/c++-Editor-Content Assist-Advanced 将未勾选的全部勾选

  6. PTA 7-2 符号配对

    直接用栈模拟即可,数组可做,但因为这节数据结构是栈,为了期末考试还是手写一下栈的操作,值得注意的是,这道题用gets函数在PTA上会编译错误,用scanf("%[^\n]", st ...

  7. hdu 5533

    Dancing Stars on Me Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Ot ...

  8. x200 xp 驱动下载

    http://support.lenovo.com/en_US/downloads/detail.page?&LegacyDocID=MIGR-70602

  9. Java EE - Servlet 3.0 和 Spring MVC

    Table of Contents 前言 基于 Java 的配置 ServletContainerInitializer 动态配置 DispatcherServlet 和 ContextLoaderL ...

  10. day01_11.break和continue

    1.continue 下一个(用next更加形象一点)整体的循环没有被破坏掉,而是跳到下一个循环单位中 <?php for($i=1;$i<=10;$i++){ if($i==4){ co ...