145. 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?

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode* root) {
vector<int> results;
if(root==NULL) return results;
if(root->right==NULL&&root->left==NULL)
{
results.push_back(root->val);
return results;
}
vector<int> lefts=postorderTraversal(root->left);
for (std::vector<int>::iterator i = lefts.begin(); i != lefts.end(); ++i)
{
results.push_back(*i);
}
vector<int> rights=postorderTraversal(root->right);
for (std::vector<int>::iterator i = rights.begin(); i != rights.end(); ++i)
{
results.push_back(*i);
}
results.push_back(root->val);
return results;
}
};