习题6-3 二叉树重建 UVa536

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1.题目描述：点击打开链接

2.解题思路：一开始的思路是先建立二叉树，再后序遍历，结果不是TLE就是MLE ==，最后省略了建树的过程，直接由先序遍历串，中序遍历串进行后序遍历。

3.代码：

#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<cassert>#include<cstdlib>#include<cstdio>#include<ctime>#include<cmath>#include<cstring>#include<functional>using namespace std;string pre_order, in_order, post_order;int n;bool solve(const string&pre,const string&in){if (pre.length() == 0) return false;//空串if (pre.length() == 1){ cout << pre; return true;}//只有一个字符时输出int p = in.find(pre[0]);string pre1 = pre.substr(1, p);    //直接进行后序遍历string in1 = in.substr(0, p);solve(pre1, in1);pre1 = pre.substr(p + 1, pre.length() - p - 1);in1 = in.substr(p + 1, in.length() - p - 1);solve(pre1, in1);cout << pre[0];}int main(){while (cin>>pre_order>>in_order){post_order.clear();n = pre_order.length();solve(pre_order,in_order);cout << "\n";}return 0;}

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