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 $N$ (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to $N$). Then 2 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<fstream>
#include<vector>
#include<stack>
#include<string>
using namespace std;

struct Order{
	string order;
	int number;
};
void OrderNum(const vector<Order> &Order1,vector<int> &Preorder,vector<int> &Inorder)
{
	stack<int> s;
	for(size_t i=0;i<Order1.size();++i){
		if(Order1[i].order =="Push"){
			s.push(Order1[i].number);
			Preorder.push_back(Order1[i].number );
		}else if(Order1[i].order =="Pop"){
			Inorder.push_back(s.top());
			s.pop();
		}
	}
}

int * Trannsform(vector<int> order)
{
	int* ret = new int(order.size());
	for(int i=0;i<order.size();++i){
		ret[i] = order[i];
	}
	return ret;
}
//**************************************************************************
struct TreeNode
{
	struct TreeNode* left;
	struct TreeNode* right;
	int  elem;
};
int Count = 0;
void BinaryTreeFromOrderings(int* inorder, int* preorder, int length)
{
	if (length == 0)
	{
		//cout<<"invalid length";
		return;
	}
	int node_value = *preorder;
	int rootIndex = 0;
	for (; rootIndex < length; rootIndex++)
	{
		if (inorder[rootIndex] == *preorder)
			break;
	}
	//Left
	BinaryTreeFromOrderings(inorder, preorder + 1, rootIndex);
	//Right
	BinaryTreeFromOrderings(inorder + rootIndex + 1, preorder + rootIndex + 1, length - (rootIndex + 1));
	Count++;
	if(Count==1)
		cout << node_value;
	else
		cout<<" "<< node_value;
	return;
}
int main()
{
	//ifstream inFile("C:\\Users\\DELL\\Desktop\\in.txt", ios::in);
	vector<int> Preorder, Inorder;
	vector<Order> Order1;
	int N;
	cin >> N;
	for(int i=0;i<2*N;++i){
		Order o;
		cin >> o.order;
		if(o.order =="Push"){
			cin >> o.number;
		}
		Order1.push_back(o);
	}
	OrderNum(Order1, Preorder, Inorder);
	int* Preorder1 = Trannsform(Preorder);
	int* Inorder1 = Trannsform(Inorder);
	BinaryTreeFromOrderings(Inorder1,Preorder1,N);
	//inFile.close();
	system("pause");
	return 0;
}