03-树2 Tree Traversals Again

这题是第二次做了，两次都不是独立完成，不过我发现我第一次参考的程序，也是参考老师（陈越）的范例做出来的。我对老师给的做了小幅修改，因为我不想有全局变量的的存在，所以我多传了三个参数进去。正序遍历每次都是从1到N吗？看题目我认为应该是，结果我错了，我是对比正确的程序一点点修改才发现的，不容易啊。下面是题目及程序

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

 typedef struct
 {
     int * a;
     int top;
 }SeqStack;

 void push(SeqStack * pS, int X);
 int pop(SeqStack * pS);
 void solve(int * pre, int * in, int * post, int preL, int inL, int postL, int n);

 int main()
 {
 //    freopen("in.txt", "r", stdin); // for test
     int i, N;
     scanf("%d", &N);

     SeqStack S;
     S.a = (int *)malloc(N * sizeof(int));
     S.top = -;
     int pre[N], in[N], post[N];

     char chars[];
     char * str = chars;
     int X, pre_index, in_index;
     pre_index = in_index = ;
     for(i = ; i <  * N; i++)
     {
         scanf("%s", str);
         if(strcmp(str, "Push") == )
         {
             scanf("%d", &X);
             pre[pre_index++] = X;
             push(&S, X);
         }
         else
             in[in_index++] = pop(&S);
     }

     solve(pre, in, post, , , , N);
     for(i = ; i < N; i++)
     {
         printf("%d", post[i]);
         if(i < N - )
             printf(" ");
         else
             printf("\n");
     }
 //    fclose(stdin); // for test
     return ;
 }

 void push(SeqStack * pS, int X)
 {
     pS->a[++(pS->top)] = X;
 }

 int pop(SeqStack * pS)
 {
     return pS->a[pS->top--];
 }

 void solve(int * pre, int * in, int * post, int preL, int inL, int postL, int n)
 {
     int i, root, L, R;

     if(n == )
         return;
     if(n == )
     {
         post[postL] = pre[preL];
         return;
     }
     root = pre[preL];
     post[postL + n - ] = root;
     for(i = ; i < n; i++)
         if(in[inL + i] == root)
             break;
     L = i;
     R = n - L - ;
     solve(pre, in, post, preL + , inL, postL, L);
     solve(pre, in, post, preL + L + , inL + L + , postL + L, R);
 }

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 keys 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 (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1