Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
   1
    \
     2
    /
   3
return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

#include 
#include 
#include 
 
using namespace std;
 
typedef struct TreeNode 
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
}TreeNode;
 
class Solution {
public:
  vector postorderTraversal(TreeNode* root) {
    TreeNode *p = NULL, *q = NULL;
    vector res;
    stack st;
    p = root;
    do
    {
      //一直向左侧走
      while (p != NULL)
      {
        st.push(p);
        p = p->left;
      }
      q = NULL;
      while (!st.empty())
      {
        p = st.top();
        st.pop();
        //q记录的不是p的左子树就是p的右子树
        //q若是p的右子树，说明p已经是第二次出栈了
        if (p->right == q)
        {
          res.push_back(p->val);
          q = p;
        }
        //q不是p的右子树，说明p的右子树还未访问
        //那q就应该再次入栈，开始访问右子树
        else
        {
          st.push(p);
          p = p->right;
          break;
        }
      }
    } while (!st.empty());
    return res;
  }
};
 
//测试用例
int main()
{
  TreeNode *pa, *pb, *pc, *pd;
 
  pa = new TreeNode(1);
  pb = new TreeNode(2);
  pc = new TreeNode(3);
  pd = new TreeNode(4);
 
  pa->left = pb;
  pa->right = pc;
  pb->right = pd;
  Solution so;
  vector res = so.postorderTraversal(pa);
 
  for each (int var in res)
  {
    cout << var << " ";
  }
}

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