03-Tree 2 Tree Traversals Again, 03-traversals

03-Tree 2 Tree Traversals Again, 03-traversals

I did this for the second time, but neither was done independently. However, I found that the program I used for my first reference was also developed based on the example of my instructor (Chen Yue. I made a slight modification to the instructor, because I don't want to have a global variable, So I uploaded three more parameters. Does a forward traversal always start from 1 to N? I think the question should be, but I am wrong. It is not easy for me to make a few changes to the correct program. Below are the questions and procedures

 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 test18     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         else40             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         else50             printf("\n");51     }52 //    fclose(stdin); // for test53     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 (); push (4); 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:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1