6-3 二叉树的重建 uva536

#include<bits/stdc++.h>
using namespace std;

int ri[1000];int le[1000];
char xian[30],zhong[30];vector<char>ans;

int built(int x1,int y1,int x2,int y2)
{
    if(x2>y2||x1>y1)return 0;

    char root=xian[x1];
     ans.push_back(root);
    int p=x2;
    while(zhong[p]!=root)p++;
    int cnt=p-x2;

     ri[root]=built(x1+cnt+1,y1,x2+cnt+1,y2);
     le[root]=built(x1+1,x1+cnt,x2,x2+cnt-1);
    return root;


}

void dfs(int root)
{

    if(le[root]){dfs(le[root]);}
    if(ri[root]){dfs(ri[root]);}
    printf("%c",root);
    return;
}


int main()
{


    while(scanf("%s %s",xian+1,zhong+1)==2)
    {  ans.clear();
       
        int n=strlen(xian+1);
 
       int root=built(1,n,1,n);
    
       //for(int i=ans.size()-1;i>=0;i--)
      //  cout<<ans[i];
      dfs(root);
       cout<<endl;



    }






    return 0;
}