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 = -1;
    int pre[N], in[N], post[N];
    
    char chars[5];
    char * str = chars;
    int X, pre_index, in_index;
    pre_index = in_index = 0;
    for(i = 0; i < 2 * N; i++)
    {
        scanf("%s", str);
        if(strcmp(str, "Push") == 0)
        {
            scanf("%d", &X);
            pre[pre_index++] = X;
            push(&S, X);
        }
        else
            in[in_index++] = pop(&S);
    }
    
    solve(pre, in, post, 0, 0, 0, N);
    for(i = 0; i < N; i++)
    {
        printf("%d", post[i]);
        if(i < N - 1)
            printf(" ");
        else
            printf("
");
    }
//    fclose(stdin); // for test
    return 0;
}

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 == 0)
        return;
    if(n == 1)
    {
        post[postL] = pre[preL];
        return;
    }
    root = pre[preL];
    post[postL + n - 1] = root;
    for(i = 0; i < n; i++)
        if(in[inL + i] == root)
            break;
    L = i;
    R = n - L - 1;
    solve(pre, in, post, preL + 1, inL, postL, L);
    solve(pre, in, post, preL + L + 1, inL + L + 1, 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