Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,3,2].

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / 
 2   3
    /
   4
    
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}"

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void inorder(TreeNode* root, vector<int> &vet)
    {
        if(!root)
            return;
        inorder(root->left,vet);
        vet.push_back(root->val);
        inorder(root->right,vet);
    }
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> ve;
        if(!root)
            return ve;
        inorder(root, ve);
        return ve;
        
    }
};