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：

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 
 5 void get_post(int *pre, int *in, int *pt, int num);
 6 
 7 int main()
 8 {
 9     int num;    //the number of nodes
10     int data;
11     int i;
12     char opr[6];
13     int *pre = NULL;    //preorder
14     int *in = NULL;     //inorder
15     int *post = NULL;   //postorder
16     int *tmp = NULL;    //a temporary array, used as a stack
17     int *pe;
18     int *pi;
19     int *pt;
20     int *pm;
21     scanf ("%d", &num);
22     pre = (int *)malloc(num*sizeof(int));
23     in = (int *)malloc(num*sizeof(int));
24     post = (int *)malloc(num*sizeof(int));
25     tmp = (int *)malloc(num*sizeof(int));
26     if (!pre||!in||!post||!tmp)
27     {
28         printf ("fail to allocate space!\n");
29         exit (1);
30     }
31     pe = pre;
32     pi = in;
33     pt = post;
34     pm = tmp;
35     for (i=0; i<2*num; i++)
36     {
37         scanf ("%s", opr);
38         if (strcmp("Push", opr)==0)
39         {//the operation is push
40             scanf ("%d", &data);
41             *pe++ = data;
42             *pm++ = data;
43         }
44         else
45         {//the operation is pop
46             *pi++ = *--pm;
47         }
48     }
49     //get the postorder traversal sequence of binary tree;
50     pt += num-1;
51     get_post(pre, in, pt, num);
52     for (i=0; i<num; i++)
53     {
54         printf ("%d ", post[i]);
55     }
56     return 0;
57 }
58 
59 void get_post(int *pre, int *in, int *pt, int num)
60 {
61     int i;
62     if (num==0) return;
63     for (i=0; i<num; i++)
64     {
65         if (in[i]==pre[0])
66             break;
67     }
68     *pt = pre[0];
69     get_post(pre+1, in, pt-(num-i), i);        //get the postorder traversal sequence of left-child
70     get_post(pre+i+1, in+i+1, pt-1, num-i-1);  //get the postorder traversal sequence of right-child
71 }

 

转载于:https://www.cnblogs.com/slfblog/p/4719304.html