uva11234 Expressions

题意:给定一个序列，小写字母代表的是数字，大写字母代表的是计算符号，这个序列以后序表达式输出，让你求出这个序列的广搜的逆序列

解题思路：使用栈构建表达式树，然后广搜求解

解题代码：

// File Name: uva11234.c
// Author: darkdream
// Created Time: 2013年05月16日 星期四 19时18分02秒

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#include<ctype.h>
struct node
{
  char c;
  struct node *right ,*left;
};
int main(){

   //freopen("/home/plac/problem/input.txt","r",stdin);
   //freopen("/home/plac/problem/output.txt","w",stdout);
   int t;
   scanf("%d\n",&t);
   while(t--)
   {
     char str[10005];
     gets(str);
     int k = strlen(str);
     struct node nodes[10005];
     struct node *hehe[10005],*head,*p;
     for(int i = 0; i < k ;i ++)
     {
        nodes[i].c = str[i];
        nodes[i].right = nodes[i].left  = NULL;
     }
     int j  = -1 ;
     for(int i = 0 ;i < k; i ++)
     {
           j++;
           hehe[j] = &nodes[i];
        if(isupper(hehe[j]->c))
        {
            hehe[j]->right = hehe[j-2];
            hehe[j]->left = hehe[j-1];
            hehe[j-2] = hehe[j];
            hehe[j-1] = NULL;
            hehe[j] = NULL;
           j = j -2;
        }
     }
    int up = 1, low = 0;
    while(low < up)
    {
       if(hehe[low]->right != NULL)
        hehe[up++] = hehe[low]->right;
       if(hehe[low]->left != NULL)
        hehe[up++] = hehe[low]->left;
       low++;
    }
    for(int i = up-1; i >= 0 ;i--)
        printf("%c",hehe[i]->c);
    printf("\n");
     
   }
return 0 ;
}