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.

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

3 4 2 6 5 1

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           using namespace std;struct Node{ int data; Node * left; Node * right; Node():data(0),left(NULL),right(NULL) {}; Node(int x):data(x),left(NULL),right(NULL) {};};void Post(Node *T,vector
           
            * o){ if(T->left) Post(T->left,o); if(T->right) Post(T->right,o); (*o).push_back(T->data);}int main(){ int n,num; cin>>n; string str; Node *lastpop=NULL,*root = NULL; stack
            
             sta; for(int i=0; i<2*n; ++i) { cin>>str; if(str=="Push") { cin>>num; if(sta.empty()&&lastpop==NULL) //上次没弹出，这次压入，根节点 { root = new Node(num); sta.push(root); } else if(lastpop) //上次有弹出，这次压入为右节点 { lastpop->right = new Node(num); sta.push(lastpop->right); } else//栈不为空且上次未弹出为压入栈顶元素左子树 { sta.top()->left = new Node(num); sta.push(sta.top()->left); } lastpop = NULL; } else { lastpop = sta.top(); sta.pop(); } } vector
             
              outt;//输出序列 Post(root,&outt); for_each(outt.begin(),outt.end(),[&](int a) { static int count = 0; cout<