5-5 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 (≤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

http://pta.patest.cn/pta/test/16/exam/4/question/667

// 5-5 Tree Traversals Again   (25分)
// http://pta.patest.cn/pta/test/16/exam/4/question/667
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <iterator>
using namespace std ;

#define INF  999999
#define N 33

int n; 

typedef struct node{
	int data ;
	struct node* left ;
	struct node* right ;
	node(int _data)
	{
		data = _data ;
		left = NULL ;
		right = NULL ;
	}
}Bnode ;

int pre[N] ;
int in[N] ;

Bnode* createTree(int preL , int preR ,int inL , int inR)
{
	if(inL > inR)
		return NULL ;
	int data = pre[preL] ;
	Bnode* r = new node(data) ;
	int pos = 0 ;
	while(in[pos++] != data);
	pos -- ;
	// 1 2 3 4 5 6 pre
	// 3 2 4 1 6 5 in 
	r->left = createTree(preL + 1, preL + pos - inL  , inL , pos - 1) ;
	r->right = createTree(preL + pos - inL + 1, preR  , pos + 1, inR) ;
	return r ;
}

vector<int> post ;

void postOrder(Bnode* root)
{
	if(root != NULL)
	{
		postOrder(root->left) ;
		postOrder(root->right) ;
		post.push_back(root->data) ;
	}
}

int main()
{
	//freopen("in.txt","r",stdin);
	scanf("%d",&n) ;
	stack<int> st ;
	int i ,val ;
	char s[10] ;
	int prei = 0 ;
	int ini = 0 ;
	for(i = 0 ; i < 2*n; i ++)
	{
		scanf("%s" , s) ;
		if(s[1] == 'u')
		{
			scanf("%d", &val);
			pre[prei++] = val;
			st.push(val);
		}else{
			int pval = st.top();
			st.pop() ;
			in[ini++] = pval ;
		}
	}
	Bnode* root = createTree(0 , n-1 , 0 , n-1) ;
	postOrder(root) ;
	printf("%d",post[0]) ;
	for(i = 1 ;i < n ; i ++)
	{
		printf(" %d" ,post[i]);
	}
	printf("\n");
	return 0 ; 
}