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.

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

題意：就是根據pop實現的是中序遍歷，按照壓入的棧中的順序是先序遍歷，最後按照後序遍歷打印出來

解題思路：利用之前的先序中序轉為後序的模板，然後在主函式中將資料稍做處理就行。pop的是中序遍歷，push是先序遍歷，因為題目沒有說明下標不重複，所以需要用value陣列儲存輸入的值（不這樣考慮只有9分），根據字串的長度可以判斷是push還是pop，具體的見程式碼

#include<iostream>
#include<cstdio>
#include<vector>
#include<stack>
#include<cstring>
using namespace std;

vector<int>in,pre,post,value;
int n;

//套模板
void postTravel(int root,int start,int end)
{
	if(start>end)
	return ;
	int i=start;
	while(i<end&&in[i]!=pre[root]) i++;
	postTravel(root+1,start,i-1);
	postTravel(root+1+i-start,i+1,end); 
	post.push_back(value[pre[root]]);
}

int main(void)
{
	scanf("%d",&n);
	stack<int>s;
	char input[7];
	int index=0,a;
	for(int i=1;i<=2*n;i++)
	{
		scanf("%s",input);
		if(strlen(input)==4)
		{
            scanf("%d",&a);
            value.push_back(a);      
		    pre.push_back(index);
			s.push(index++);
		}else{
			in.push_back(s.top());
			s.pop();
		}	
	}
	postTravel(0,0,n-1);
	for(int i=0;i<n;i++)
	printf("%d%s",post[i],i==n-1?"\n":" ");
	return 0;
}