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

 1 #include <stdio.h>

 2 #include <stdlib.h>

 3 #include <string.h>

 4 

 5 typedef struct

 6 {

 7     int * a;

 8     int top;

 9 }SeqStack;

10 

11 void push(SeqStack * pS, int X);

12 int pop(SeqStack * pS);

13 void solve(int * pre, int * in, int * post, int preL, int inL, int postL, int n);

14 

15 int main()

16 {

17 //    freopen("in.txt", "r", stdin); // for test

18     int i, N;

19     scanf("%d", &N);

20     

21     SeqStack S;

22     S.a = (int *)malloc(N * sizeof(int));

23     S.top = -1;

24     int pre[N], in[N], post[N];

25     

26     char chars[5];

27     char * str = chars;

28     int X, pre_index, in_index;

29     pre_index = in_index = 0;

30     for(i = 0; i < 2 * N; i++)

31     {

32         scanf("%s", str);

33         if(strcmp(str, "Push") == 0)

34         {

35             scanf("%d", &X);

36             pre[pre_index++] = X;

37             push(&S, X);

38         }

39         else

40             in[in_index++] = pop(&S);

41     }

42     

43     solve(pre, in, post, 0, 0, 0, N);

44     for(i = 0; i < N; i++)

45     {

46         printf("%d", post[i]);

47         if(i < N - 1)

48             printf(" ");

49         else

50             printf("\n");

51     }

52 //    fclose(stdin); // for test

53     return 0;

54 }

55 

56 void push(SeqStack * pS, int X)

57 {

58     pS->a[++(pS->top)] = X;

59 }

60 

61 int pop(SeqStack * pS)

62 {

63     return pS->a[pS->top--];

64 }

65 

66 void solve(int * pre, int * in, int * post, int preL, int inL, int postL, int n)

67 {

68     int i, root, L, R;

69     

70     if(n == 0)

71         return;

72     if(n == 1)

73     {

74         post[postL] = pre[preL];

75         return;

76     }

77     root = pre[preL];

78     post[postL + n - 1] = root;

79     for(i = 0; i < n; i++)

80         if(in[inL + i] == root)

81             break;

82     L = i;

83     R = n - L - 1;

84     solve(pre, in, post, preL + 1, inL, postL, L);

85     solve(pre, in, post, preL + L + 1, inL + L + 1, postL + L, R);

86 }

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