题目：http://pta.patest.cn/pta/test/16/exam/4/question/667
PTA - Data Structures and Algorithms (English) - 5-5
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.
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 //即后序遍历树的结果解法转自：http://www.cnblogs.com/clevercong/p/4177802.html
题目分析：
动态的“添加结点”并遍历过程，不需要建树。
借助“栈”进行树的后续遍历；( times：可由此栈操作判断中当前结点有左子树还是右子树中 )
每次PUSH，times = 1；// PUSH：可看作访问左孩子
每次POP检查栈顶记录的times：
- 如果是1，弹出来变成2压回栈；// POP：可看作由左孩子转到访问右孩子
- 如果是2，则弹出，放入存放结果的vector中，重复这一过程，直到栈顶times为1。// POP：可看作访问根节点
所有PUSH与POP操作执行完毕时，输出vector内的数据和stack中的数据。 注意要处理最后的空格。
注：此方法可推广到n叉树，只需改变times的值即可。
……
代码：
#include <iostream>
#include <stack>
#include <vector>
using namespace std;
typedef struct node
{
int data;
int times;
node(int d, int t)
:data(d), times(t)
{};
} Node;
int main()
{
int n;
cin >> n;
string cmd; //!PUSH or POP
int x;
stack<Node> sta; //!栈：用来实现后序遍历
vector<int> vec; //!存放结果
for(int i=0; i<2*n; i++)
{
cin >> cmd;
if(cmd == "Push")
{
cin >> x;
sta.push(Node(x, 1));
}
if(cmd == "Pop")
{
Node node = sta.top();
sta.pop();
if(node.times == 1) //!times==1：弹出来变成2压回栈
{
node.times = 2;
sta.push(node);
}
else if(node.times == 2) //!times==2：弹出，放入存放结果的vector中
{
vec.push_back(node.data);
while(sta.top().times == 2) //!重复这一过程，直到栈顶的times==1
{
vec.push_back(sta.top().data);
sta.pop();
}
if (sta.size() != 0) //!操作结束后栈还不为空，使栈顶的times=2
{
sta.top().times=2;
}
}
}
}
for(unsigned int i=0; i<vec.size(); i++)
{
cout << vec[i]<< " ";
}
while(sta.size() != 0) //!PUSH/POP结束后栈中还有元素，直接pop()即可
{
cout << sta.top().data;
sta.pop();
if(sta.size() != 0)
cout << " ";
}
return 0;
}