PTA Tree Traversals Again

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

代码：

#include<iostream>
#include<stack>
#include<string>
#include<stdio.h>
#include<malloc.h>
using namespace std;
typedef struct TreeNode* Tree;
struct TreeNode {
	string data;
	Tree left;   // 左子树 
	Tree right;  // 右子树 
};
// 初始化一个树结点 
Tree create() {
	Tree T;
	T = (Tree)malloc(sizeof(struct TreeNode));
	T->left = NULL;
	T->right = NULL;
	return T;
}

// 根据中序遍历整理出这棵树 
Tree restore(Tree T) {
	int n;
	string str;
	stack<Tree> s;
	Tree node = T;
	bool flag = false;
	string value;
	scanf("%d\n", &n);
	getline(cin, str);
	value = str.substr(5);  //
	node->data = value;
	// 根结点入栈 
	s.push(node);
	for (int i = 1; i < 2 * n; i++) {
		getline(cin, str);
		if (str == "Pop") {// 如果是 pop 操作
			node = s.top();
			s.pop();
		}
		else {   // push
			value = str.substr(5);  // 从第五个开始截取
			Tree tmp = create();
			tmp->data = value;
			if (!node->left) {// 如果左儿子空，新结点就是左儿子 
				node->left = tmp;
				node = node->left;
			}
			else if (!node->right) {  // 如果右儿子空，新结点就是右儿子 
				node->right = tmp;
				node = node->right;
			}
			s.push(tmp);
		}
	}
	return T;
}

// 后序递归遍历
void bl(Tree T, bool& flag) {
	if (T) {
		bl(T->left, flag);
		bl(T->right, flag);
		if (!flag)
			flag = true;
		else
			cout << " ";
		cout << T->data;
	}
}
int main() {
	Tree T;
	bool flag = false;
	string str;
	T = create();
	T = restore(T);
	bl(T, flag);
	return 0;
}

提交结果：