UVa, 10701 Pre, in and post

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本文详细介绍了如何通过给定的前序和中序遍历序列来计算二叉树的后序遍历序列，包括算法实现、输入输出处理和注意事项。

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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