03-树3 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.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer $N$ ($\le 30$) 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<iostream>
#include<string>
#include<stack>
using namespace std;
int first=1;
typedef struct Dnode
{
    int data;
    struct Dnode *up;
    struct Dnode *l;
    struct Dnode *r;
}Tree;
void postOrderPrint(Tree *p)
{
    if(p){
        postOrderPrint(p->l);
        postOrderPrint(p->r);
        if(first){
        	cout<<p->data;
        	first=0;
        }
        else cout<<' '<<p->data;
    }
}
int main()
{
    int n,num,lastOpea=0,cnt=0;
    string s;
    Tree *p=new Tree;
    p->up=p->l=p->r=NULL;
    cin>>n>>s>>p->data;
    stack<int>points;
    points.push(p->data);
    while(cnt<n-1 && cin>>s){
        if(s.compare("Push")==0){
        	cin>>num;
        	Tree *tem=new Tree;
            tem->data=num;
            tem->up=p;
            while(1){
            	if(p->data==points.top()){
	                if(!lastOpea && !p->l){
	                    p=p->l=tem;
	                    break;
	                }
	                else if(lastOpea && !p->r){
	                    p=p->r=tem;
	                    points.pop();
	                    break;
	                }
	            }
	            else p=p->up;
            }
            points.push(num);
            p->l=p->r=NULL;
            cnt++;
            lastOpea=0;
        }
        else{
        	if(lastOpea) points.pop();
        	lastOpea=1;
        }
    }
    while(p->up) p=p->up;
    postOrderPrint(p);
    delete p;

    return 0;
}