UVa, 10701 Pre, in and post

Problem F - Pre, in and post

Time Limit: 1 second

A common problem in data structures is to determine the traversal ofa binary tree. There are three classic ways to do it:

Pre-order: You must visit insequence the root, left subtree and the right subtree.
In-order: You must visitin sequence the left subtree, the root and the right subtree.
Post-order: You must visit insequence the left subtree, the right subtree and the root.

See the picture below:

The pre, in and post-order traversal are, respectively, ABCDEF, CBAEDF and CBEFDA. In this problem, you must compute the post-ordertraversal of a binary tree given its in-order and pre-order traversals.

Input

The input set consists of a positive number C ≤ 2000, thatgives the number of test cases and Clines, one for each test case. Each test case starts with a number 1≤ N≤ 52, the number of nodes in this binary tree. After, there will betwo strings S₁ and S₂ thatdescribe the pre-order and in-order traversal of the tree. Thenodes of the tree are labeled with different characters in the range a..zand A..Z. The values of N,S₁ and S₂ are separeted by a blankspace.

Output

For each input set, you should output a line containing the post-ordertransversal for the current tree.

Sample input

3
3 xYz Yxz
3 abc cba
6 ABCDEF CBAEDF

Sample output

Yzx
cba
CBEFDA

Problem setter: Sebastião Alves #include<iostream>
#include<string>
#include<vector>
#include<cstdio>
using namespace std;
///////////////////////////
class Preinpost{
private:
    string preorder;
    string inorder;
    string postorder;
    void reconstruct(string p,string i);
public:
    //void readcase(){;
    void initial(string p,string i)
    {
        preorder=p;
        inorder=i;
        postorder.clear();
    }
    void computing();
    void outresult(){cout<<postorder<<endl;}

};
void Preinpost::computing(){
    reconstruct(preorder,inorder);
}
void Preinpost::reconstruct(string preorder,string inorder){
    int si=preorder.size();
    if(si>0){
        int p=int(inorder.find(preorder[0]));
        reconstruct(preorder.substr(1,p),inorder.substr(0,p));
        reconstruct(preorder.substr(p+1,si-p-1),inorder.substr(p+1,si-p-1));
        postorder.push_back(preorder[0]);
    }
}
int main(){
   Preinpost pip;

   string p,i;
   //cin>>n;
   //getchar();
   int n,m;
   cin >> n;
   while(n--){
         //int m;
         cin>>m;
         cin>>p>>i;
         //pip.readcase();
         pip.initial(p,i);
         pip.computing();
         pip.outresult();
   }
   return 0;
}

二叉树 的基本问题 注意下 recontruct 的算法 体会下 递归的函数 的使用，再者就是 输入输出的变换。。。灵活使用 。。。表示因为输入问题wa

了两次，后来 认识到不是算法的问题 ，改了下输入的方法就ac