Tree Traversals Again

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 </center>

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer

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`

题意：

大概就是用push、pop创建一个二叉树，（先序与中序），然后用后序输出。

咋做：

刚开始用的是结构体数组来做，我丢，调试了一整天一直都是那个测试点没过 >< ,然后听别人说输入的节点的值数值可能相等，我丢..........

后来改成了链式结构，然后有个小地方注意一下就A了。

代码：

#include <string>
#include <map>
#include <cmath>
#include <limits>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <stack>
#include <queue>
#define Sci(x) scanf("%d",&x)
#define Sci2(x, y) scanf("%d%d",&x,&y)
#define Sci3(x, y, z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%I64d",&x)
#define Scl2(x, y) scanf("%I64d%I64d",&x,&y)
#define Scl3(x, y, z) scanf("%I64d%I64d%I64d",&x,&y,&z)
#define Pri(x) printf("%d\n",x)
#define Prl(x) printf("%I64d\n",x)
#define For(i,x,y) for(int i=x;i<y;i++)
#define FFor(i,x,y) for(int i=x;i>y;i--)
#define For_(i,x,y) for(int i=x;i<=y;i++)
#define FFor_(i,x,y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))
#define LL long long
#define ULL unsigned long long
#define MAXSIZE 30
#define INF 0x3f3f3f3f
const int mod = 1e9+7;
const double PI = acos(-1.0);

using namespace std;
typedef struct node
{
    struct node *left;
    struct node *right;
    int data;
} Tree,*Ptr;
Ptr init(int a)
{
    Ptr t;
    t=new Tree;
    t->left=t->right=NULL;
    t->data=a;
    return t;
}
int flag=1;
void putt(Ptr t)
{
    if(t!=NULL)
    {
        putt(t->left);
        putt(t->right);
        if(flag)
            printf("%d",t->data);
        else
            printf(" %d",t->data);
        flag=0;
    }
}
int main()
{
    int n;
    Sci(n);
    Ptr T,p;
    T=new Tree;
    T->left=T->right=NULL;
    p=T;
    int a;
    stack<Ptr>s;
    char str[6];
    s.push(T);
    while(~scanf("%s",str))
    {
        if(strcmp(str,"Push")==0)
        {
            scanf("%d",&a);
            if(p->left==NULL)
            {
                p->left=init(a);
                s.push(p->left);
                p=p->left;
            }
            else if(p->right==NULL)
            {
                p->right=init(a);
                s.push(p->right);
                p=p->right;
            }
        }
        else
        {
            p=s.top();
            s.pop();
        }
    }
    putt(T->left);//此处传递的是根的左结点？？？上面的while循环里最开始入栈的是根的左结点，在题目输入时把这个结点当作根节点来使用，所以  T->left. QAQ
    return 0;
}