【二叉樹】POJ 2255

題意很簡單，給出一棵樹的前序&中序遍曆，輸出後續遍曆。

首先我們來研究一下第一個sample：

DBACEGF        ABCDEFG

前序走訪是先是從根節點->左子樹->右子樹；而中序遍曆則是左子樹->根節點->右子樹。所以DBACEGF中D是根， ABC D
EFG 中序可以找左子樹&右子樹，此時把D放在輸出數組的最後一個，然後先遞迴處理右子樹再處理左子樹。

#include <map>#include <set>#include <list>#include <queue>#include <deque>#include <stack>#include <string>#include <time.h>#include <cstdio>#include <math.h>#include <iomanip>#include <cstdlib>#include <limits.h>#include <string.h>#include <iostream>#include <fstream>#include <algorithm>using namespace std;#define LL long long#define MIN INT_MIN#define MAX INT_MAX#define PI acos(-1.0)#define FRE freopen("input.txt","r",stdin)#define FF freopen("output.txt","w",stdout)#define N 30char pre[N];char in[N];char ans[N];int len;void gao(int s1, int s2, int t1, int t2){    int i;    if(s1 > s2)return ;    for(i = t1; i <= t2; i++){        if(in[i] == pre[s1]){            break;        }    }    ans[--len] = pre[s1];    if(s1 == s2){        return ;    }    gao(s1 + i - t1 + 1, s2, i + 1, t2);    gao(s1 + 1, s1 + i - t1, t1, i - 1);}int main(){    while(scanf("%s%s",pre,in) != EOF){        int i,j;        len = strlen(pre);        memset(ans,'\0',sizeof(ans));        gao(0,len - 1,0,len - 1);        printf("%s\n",ans);    }    return 0;}