03-樹3 Tree Traversals Again （25 point(s)）

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：總結下來就是：用stack模擬了二叉樹的先序和中序，我們要輸出該二叉樹的後序遍歷。 push就是先序，pop就是中序，我們可以建立一個stack容器來儲存。

程式碼如下：

#include <iostream>
#include <stack>
using namespace std;
const int maxn=100;
struct tree{
	int data;
	tree *l;
	tree *r;
};
int per[maxn],in[maxn];
tree* Buildtree(int pl,int pr,int il,int ir){          
	if(pl>pr){
		return NULL;
	}
	tree *root = new tree;
	root->data=per[pl]; 
	int k;
	for(k=il;k<=ir;k++){
		if(in[k]==root->data)
			break;
	}
	int lnum=k-il;
	root->l=Buildtree(pl+1,pl+lnum,il,k-1);
	root->r=Buildtree(pl+lnum+1,pr,k+1,ir);
	return root;
}
int flag=0;
void postorder(tree *root){
	if(root==NULL)
		return;
	postorder(root->l);
	postorder(root->r);
	if(!flag){
		cout<<root->data;	
		flag=1;
	}
	else
		cout<<" "<<root->data;        //末尾不能有多餘空格
}

int main()
{
	int n;
	cin>>n;
	int cnt=0;
	string str;
	int num,i=0,k=0;
	stack <int>s;            
	while(cnt<n){            //當Pop次數和要輸入的資料相等時不再輸入
		cin>>str;
		if(str=="Push"){
			cin>>num;
			per[i++]=num;
			s.push(num);
		}
		if(str=="Pop"){ 
			int res=s.top();        //將stack容器中的元素轉移到中序遍歷陣列中
			s.pop();
			in[k++]=res;
			cnt++;                    
		} 
	}
	tree *root=Buildtree(0,n-1,0,n-1);        //根據先序和中序建立二叉樹
	postorder(root);                          //後序遍歷
	return 0;	
}