MOOC Tree Traversals Again

03-树3 Tree Traversals Again (25分)
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.
[图片上传失败...(image-1897-1511628506569)]Figure 1
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N

N

N (≤
3
0

\le 30

≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N

N

N). Then 2
N

2N

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.

#include 
#include 
#include 
#define MaxSize 30

struct{
    int top;
    int data[MaxSize]; 
}stack;

int preorder[MaxSize],inorder[MaxSize];
void GetPostOrder(int preorder[],int a,int b,int inorder[],int c,int d)
{
    int i;
    for( i=c;i<=d&&preorder[a]!=inorder[i];i++);
    if(i>0) GetPostOrder(preorder,a+1,i,inorder,c,i-1);
    if(i