POJ 2255 Tree Recovery（树的遍历）

给定前序遍历和中序遍历，写出后序遍历。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <assert.h>
#include <algorithm>
#define MAX 1234567890
#define MIN -1234567890
#define eps 1e-8

using namespace std;

char bef[32];
char mid[32];
char aft[32];
char tree[1024];
bool visit[32];
int len;
int p;

void build(char root, int t)
{
        int i, j;
        tree[t] = root;
        visit[root-‘A‘] = true;
        for(i = 0; i < len; i++) if(mid[i] == root) break;
        for(j = 0; j < len; j++) if(bef[j] == root) break;
        if(i-1 >= 0 && !visit[mid[i-1]-‘A‘])
        {
                char lson = bef[j+1];
                build(lson, 2*t);
        }
        if(i+1 < len && !visit[mid[i+1]-‘A‘])
        {
                char rson;
                for(int k = j+1; k < len; k++)
                {
                        if(!visit[bef[k]-‘A‘])
                        {
                                rson = bef[k];
                                break;
                        }
                }
                build(rson, 2*t+1);
        }
}

void after(char root, int t)
{
        if(tree[2*t] != ‘\0‘) after(tree[2*t], 2*t);
        if(tree[2*t+1] != ‘\0‘) after(tree[2*t+1], 2*t+1);
        aft[p++] = root;
}

int main()
{
        #ifdef BellWind
                freopen("2255.in", "r", stdin);
        #endif // BellWind

        while (~scanf("%s %s", bef, mid))
        {
                memset(visit, 0, sizeof(visit));
                memset(tree, 0, sizeof(tree));
                len = strlen(bef);
                build(bef[0], 1);
//                int cnt = 0;
//                for(int c = 1; cnt < len; c++)
//                {
//                        if(tree[c] == ‘\0‘) cout << ‘ ‘;
//                        else {cnt++; cout << tree[c];}
//                }
//                cout << endl;
                p = 0;
                after(bef[0], 1);
                for(int c = 0; c < len; c++) cout << aft[c];
                cout << endl;
        }

        return 0;
}