非递归实现树的前序遍历

/*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?
*/
/****************************/
struct  TreeNode{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x):val(x),left(NULL),right(NULL){}
}
vector<int> postorder(TreeNode *root)
{
     stack<TreeNode *> s;
     vector<int> res;
     if(root==NULL) return res;
     s.push(root);
     while(!s.empty()){
         ListNode *temp=s.top();
         s.pop();
         res.push_back(temp->val);
         if(temp->right!=NULL) s.push(temp->right);
         if(temp->left!=NULL) s.push(temp->left);
     }
     return res;
}