LeetCode-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 sub(TreeNode *root,vector<int>*ret){
        if(root==NULL)return;
        sub(root->left,ret);
        ret->push_back(root->val);
        sub(root->right,ret);
    }
    vector<int> inorderTraversal(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> ret;
        sub(root,&ret);
        return ret;
    }
};