根据前序中序写后序（正确写法）

题目描述

      已知二叉树的前序和中序遍历，输出该二叉树的后序遍历。例如下面二叉树的前序和中序遍历为ABDC、DBAC，后序遍历为DBCA。 

           A

          / 

         /   

        B     C

       /

      /

     D

输入

包括多组测试数据。

每组1行，包含两个字符串，分别为叉树的前序和中序遍历。

输出

二叉树的后序遍历。

样例输入

ABDC DBAC
BCAD CBAD

样例输出

DBCA
CDAB

-----------------------------------------------------------------------------------------------
举个栗子
前序：ABCDEFG
中序：CBEDAFG
思想：一步一步划分，将其存入两个数组中；从第一个根开始划分；；；；
看图：
--------------------------------------------------------------------------------------------

                         left = 1         CBEDAFG            right = 7                                    

                                                    ↓   

           left  =  1      CBED    right = 4          A                               left = 6           FG      right = 7

                               ↓                                                                                          ↓

             C    B               ED                                left = 6 right = 6
             ↓                     ↓
         left = 1  right = 1    left = 3, right = 4
                                   ↓    
          
                              left = 3,  right = 3 



1.如果left < right  那么继续递归
2，如果left == right  说明它是一个叶子节点，则T->lchild = T->right = null;
3,如果 left > right   说明它是一个null,   T=null;
----------------------------------------------------------------------------------------------

 

 #include <cstdio>
#include <cstdlib>
#include <cstring>
//#define _OJ_
typedef struct tree1
{
    char data;
    struct tree1 *lchild;
    struct tree1 *rchild;
} tree1, *tree;

tree
creat_tree(tree T, int len, int left, int right, char *t1, char *t2)
{
    if(left < right) {
    int i = 1;
    T = (tree) malloc (sizeof(tree1));    T->data = t1[len];
     while (t1[len] != t2[i])
       i++;

    T->lchild = creat_tree(T->lchild, len + 1, left, i - 1, t1, t2);
    // 递归遍历左子树  left不变， i - 1；
    T->rchild = creat_tree(T->rchild, len + i + 1 - left, i + 1, right, t1, t2);
    // 递归遍历左子树   right 不变， i + 1;len + i + 1 - left,
    }//难点在于参数的传递
    else if(left == right) {
            T = (tree) malloc (sizeof(tree1));    T->data = t1[len];
            T->lchild = T->rchild = NULL;
        }

    else if(left > right)    T = NULL;

    return T;
}

void
travers(tree T)
//递归遍历一棵树
{
    if(T) {travers(T->lchild);
    travers(T->rchild);
    printf("%c", T->data);
    }
}



int main(int argc, char const *argv[]) {
#ifndef _OJ_  //ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    tree T;
    char t1[20], t2[20];
    int i, len;

    while(scanf("%s", t1) != EOF) {
    scanf("%s", t2);

    len = strlen(t1);//0号位不用
    for(i = len;i >= 0; i--) {
    t1[i + 1] = t1[i];
    t2[i + 1] = t2[i];
     }

    T = creat_tree(T, 1, 1, len, t1, t2);
    travers(T);
    printf(" ");
 }
    return 0;
}

-----------------------------------------------------------------------------------------------------------------------

通过举一反三我们可以同理可得通过中序和后序写前序

#include <cstdio>
#include <cstdlib>
#include <cstring>
//#define _OJ_
typedef struct tree1
{
    char data;
    struct tree1 *lchild;
    struct tree1 *rchild;
} tree1, *tree;

tree
creat_tree(tree T, int len, int left, int right, char *t1, char *t2)
{
    if(left < right) {
    int i = 1;
    T = (tree) malloc (sizeof(tree1));    T->data = t2[len];
     while (t2[len] != t1[i])
       i++;
    //对于右子树，ｉ ＋ 1 , right不变
    T->rchild = creat_tree(T->rchild, len - 1, i + 1, right, t1, t2);
    T->lchild = creat_tree(T->lchild, len + i - 1 - right,    left, i - 1, t1, t2);
    }//对于左子树left不变， i - 1, 特别注意  参数len + i - 1 - right

    else if(left == right) {
     T = (tree) malloc (sizeof(tree1));    T->data = t2[len];
     T->lchild = T->rchild = NULL;
    }
    else if(left > right)
        T = NULL;

    return T;
}

void
travers(tree T)
//递归遍历一棵树
{
    if(T) {
    printf("%c", T->data);
    travers(T->lchild);
    travers(T->rchild);
    }
}



int main(int argc, char const *argv[]) {
#ifndef _OJ_  //ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    tree T;
    char t1[20], t2[20];
    int i, len;

    while(scanf("%s", t1) != EOF) {
    scanf("%s", t2);

    len = strlen(t1);//0号位不用
    for(i = len;i >= 0; i--) {
    t1[i + 1] = t1[i];
    t2[i + 1] = t2[i];
     }

    T = creat_tree(T, len, 1, 7, t1, t2);
    travers(T);
    printf(" ");
 }
    return 0;
}
// 中序：CBEDAFG
// 后序：CEDBGFA
// 前序：ABCDEFG
----------------------------------------------------------------------------------------------------------------------

len + 1+ i - left;;;;;;         len - 1 +i -right;;;;;;

其实无论怎么写代码，看别人的代码也好，，
一定要有自己的想法！！！也许别人的代码很渣！！！