SOJ-2664（已知满二叉树的后序求中序遍历）

因为知道小写字母代表叶子，大写字母代表非叶子，如后序为bcA，那么中序久违bAc，这样就可以用逆波兰表示法的二叉树做，把小写字母当成操作数，大写字母当成操作符，每有一个操作符就做一个操作。

/******************************************************************************************************
 ** Copyright (C) 2011.07.01-2013.07.01
 ** Author: famousDT <[email protected]>
 ** Edit date: 2011-12-07
******************************************************************************************************/
#include <stdio.h>
#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll
#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10
#include <vector>
#include <queue>
#include <map>
#include <time.h>
#include <set>
#include <list>
#include <stack> 
#include <string>
#include <iostream>
#include <fstream>
#include <assert.h>
#include <bitset>
#include <iterator>//C++Primer
#include <string.h>//memcpy(to,from,count
#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprll
#include <algorithm>
using namespace std;

//typedef long long int ll;

#define MY_PI acos(-1)
#define MY_MAX(a, b) ((a) > (b) ? (a) : (b))
#define MY_MIN(a, b) ((a) < (b) ? (a) : (b))
#define MY_MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))
#define MY_ABS(a) (((a) >= 0) ? (a) : (-(a)))
#define MY_INT_MAX 0x7fffffff

/*==========================================================*\
| 
\*==========================================================*/
struct node
{
	char data;
	struct node *left;
	struct node *right;
} wo[1000];
void inorder(struct node *x)
{
	if (x != NULL) {
		inorder(x->left);
		printf("%c", x->data);
		inorder(x->right);
	}
}
int main()
{
	int cases;
	char s[1000];
	scanf("%d", &cases);
	getchar();
	while (cases--) {
		stack<struct node *> st;
		gets(s);
		int i, len = strlen(s);
		for (i = 0; i < len; ++i) {
			wo[i].data = s[i];
			wo[i].left = NULL;
			wo[i].right = NULL;
		}
		for (i = 0; i < len; ++i) {
			if (islower(wo[i].data)) {
				st.push(&wo[i]);
			} else {
				struct node *a, *b;
				a = st.top();
				st.pop();
				b = st.top();
				st.pop();
				wo[i].left = b;
				wo[i].right = a;
				st.push(&wo[i]);
			}
		}
		struct node *x = st.top();
		inorder(x);
		printf("\n");
	}
	return 0;
}