03-3. Tree Traversals Again (25)

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

#include <stdio.h>
#define MaxSize 35
#define EMPTY_TOS (-1)
#define PUSH 1
#define POP  2

typedef struct stack_record
{
    int top_of_stack;
    int stack_array[MaxSize];
} Stack;
void init_stack(Stack *S)
{
    S->top_of_stack=EMPTY_TOS;
}
int
is_empty(Stack *S)
{
    return (S->top_of_stack==EMPTY_TOS);
}

int
is_full(Stack *S)
{
    return (S->top_of_stack==MaxSize-1);

}

void
push(int x,Stack *S)
{
    if(is_full(S))
     return;
    else
     S->stack_array[++S->top_of_stack]=x;
}
int
top(Stack *S){
    if(!is_empty(S))
      return S->stack_array[S->top_of_stack];
}
int
pop(Stack *S)
{
    if(is_empty(S))
        return;
    else
        return S->stack_array[S->top_of_stack--];

}

int
main ()
{
    Stack S1,S2;
    int OpArray[2*MaxSize+1][2]={-1};
    char tmpstr[30],tmp[30],tmp1[30];
    init_stack(&S1);
    init_stack(&S2);
    int i,NodeNum,tmppop,tmppop1;
    int PrintTag=0;

    scanf("%d",&NodeNum);
    getchar();
    for (i=0;i<2*NodeNum;i++){
        gets(tmpstr);
        if(tmpstr[1]==‘u‘){
            sscanf(tmpstr,"%s%s",tmp,tmp1);
            OpArray[i][0]=PUSH;
            OpArray[i][1]=atoi(tmp1);
        }else {

            OpArray[i][0]=POP;

        }        

    }

    for (i=0;i<2*NodeNum;i++){

        if(OpArray[i][0]==PUSH)
            push(OpArray[i][1],&S1);

        else{
            if(OpArray[i+1][0]==PUSH){

               if(top(&S1)==-1){
                   push(OpArray[i+1][1],&S2);

               }else{

                   push(-1,&S2);
                   push(pop(&S1),&S2);
                   push(-1,&S1);
                   push(OpArray[i+1][1],&S2);

               }
               i++;

            } else{
              tmppop=pop(&S1);
              if(tmppop==-1){
                  while((tmppop1=pop(&S2))!=-1)
                     printf(" %d",tmppop1);
              }else{
                    if (!PrintTag){
                      printf("%d",tmppop);
                      PrintTag++;
                  }else
                    printf(" %d",tmppop);

              }

            }

        }

    }

    return 0;
}