PAT Tree Traversal Again

　　This is my very first.

　　哈哈，闲话少说，转入正题。

　　学完数据结构后，老是感觉学得不扎实，于是到MOOC网上去刷题。结果在Tree Traversal Again 这道题上卡住了。题目是这么说的：

     An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree(with the key numbered from 1 to 6) is traversed, the stack operations are: Push(1), Push(2), Push(3), Pop(), Pop(), Push(4), Pop(), Pop(), Push(5), Push(6), Pop(), Pop(). Then a unique binary tree can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of the tree. （关于题目的更详细的信息看这里http://www.patest.cn/contests/mooc-ds2015spring/03-%E6%A0%912）

　　 当时没大弄明白这道题目的意思，几番纠结之后，我看了一下陈越老师的讲课视频，恍然大悟。如果给出一棵二叉树的先序遍历和中序遍历结果，那么这棵二叉树就唯一确定了。题目中给出的不就是先序遍历和中序遍历的结果吗？把入栈的过程记录下来就是先序遍历的结果，把出栈的过程记录下来就是中序遍历的结果。突然又想起来这种问题在严蔚敏老师的那本数据结构上讲过，时间一长，我竟然给忘了！翻书，果然找到了已知先序序列和中序序列如何求二叉树的方法。不过我们这里没有必要把二叉树求出来再进行后续遍历，只需按照后序遍历的顺序把元素放到数组的相应位置就可以了。下面是实现的代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void get_post(int *pre, int *in, int *pt, int num);

int main()
{
    int num;    //the number of nodes
    int data;
    int i;
    char opr[6];
    int *pre = NULL;    //preorder
    int *in = NULL;     //inorder
    int *post = NULL;   //postorder
    int *tmp = NULL;    //a temporary array, used as a stack
    int *pe;
    int *pi;
    int *pt;
    int *pm;
    scanf ("%d", &num);
    pre = (int *)malloc(num*sizeof(int));
    in = (int *)malloc(num*sizeof(int));
    post = (int *)malloc(num*sizeof(int));
    tmp = (int *)malloc(num*sizeof(int));
    if (!pre||!in||!post||!tmp)
    {
        printf ("fail to allocate space!\n");
        exit (1);
    }
    pe = pre;
    pi = in;
    pt = post;
    pm = tmp;
    for (i=0; i<2*num; i++)
    {
        scanf ("%s", opr);
        if (strcmp("Push", opr)==0)
        {//the operation is push
            scanf ("%d", &data);
            *pe++ = data;
            *pm++ = data;
        }
        else
        {//the operation is pop
            *pi++ = *--pm;
        }
    }
    //get the postorder traversal sequence of binary tree;
    pt += num-1;
    get_post(pre, in, pt, num);
    for (i=0; i<num; i++)
    {
        printf ("%d ", post[i]);
    }
    return 0;
}

void get_post(int *pre, int *in, int *pt, int num)
{
    int i;
    if (num==0) return;
    for (i=0; i<num; i++)
    {
        if (in[i]==pre[0])
            break;
    }
    *pt = pre[0];
    get_post(pre+1, in, pt-(num-i), i);        //get the postorder traversal sequence of left-child
    get_post(pre+i+1, in+i+1, pt-1, num-i-1);  //get the postorder traversal sequence of right-child
}