1086 Tree Traversals Again (25分)--PAT甲级真题

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

分析：
其实还是给出前序和中序序列，求后序遍历序列；只是题目中搞了点花样，用压栈，出栈来给出前序和中序序列；其中压栈序列，就是前序序列，出栈序列，就是中序序列；
有了前序和中序之后，进行一次后序遍历就可以得到后序序列；

#include
#include
#include 
using namespace std;  
int preorder[31], inorder[31];
int pre_cnt = 0, in_cnt = 0, loc = 0;
stack<int>Sta; 
bool firstprint = false;  
void build(int s1, int e1, int s2, int e2){   
	int rootindex;
	for (int i = s2; i <= e2; i++){
		if (inorder[i] == preorder[s1]){
			rootindex = i;
			break;
		}
	} 
	if (rootindex != s2)
		build(s1 + 1, s1 + rootindex - s2, s2, rootindex - 1);
	if (rootindex != e2)
		build(s1 + rootindex - s2 + 1, e1, rootindex + 1, e2); 
	firstprint == false ? firstprint = true : printf(" ");
	printf("%d", preorder[s1]); 
}   
int main(){
	int N, d;
	scanf("%d", &N); 
	for (int i = 0; i < 2 * N; i++){
		string op; 
		cin >> op;
		if (op == "Push"){
			scanf("%d", &d); 
			Sta.push(d);
			preorder[pre_cnt++] = d;
		}
		else{
			d = Sta.top();
			Sta.pop();
			inorder[in_cnt++] = d;
		}
	} 
	build(0, N - 1, 0, N - 1); 
	return 0;
}