Binary Tree Traversals HDU

A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2.

In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.

In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.

In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.

Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.

InputThe input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and inorder sequence. You can assume they are always correspond to a exclusive binary tree.
OutputFor each test case print a single line specifying the corresponding postorder sequence.
Sample Input

9
1 2 4 7 3 5 8 9 6
4 7 2 1 8 5 9 3 6

Sample Output

7 4 2 8 9 5 6 3 1

树的重建，其实也是套路（rec()是书上给的代码，不太好理解）！
需要注意的是vector，有时需要清空（根据题目来定）！
因为这是颗不完整的树，如果将后序存在数组里面，可能会爆掉！所以直接打印！

#include<cstdio>
#include<cstring>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> pre,in;
int n,pos,cnt;
void rec(int l,int r)
{   
    if(l>=r) return;
    int root=pre[pos++];
    int m=distance(in.begin(),find(in.begin(),in.end(),root));
    rec(l,m);
    rec(m+1,r);
    cout<<root;                                 //长数组的输出，最好直接打印出来，不存！ 
    cnt++;
    if(cnt!=n) cout<<" "; 
    if(cnt==n) cout<<endl;    
}
int main()
{   int k;
    while(cin>>n){
          pos=0,cnt=0;
       for(int i=0;i<n;i++)
       {   
           cin>>k;
           pre.push_back(k);
       }
       for(int i=0;i<n;i++)
       {
              cin>>k;
              in.push_back(k);
       }
       rec(0,pre.size()); 
       pre.clear();                      //vector在外面声明的话，如果要多次用，就要清空； 
       in.clear();                       
    }
    return 0;
}