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 do it iteratively?

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

/** * Definition for binary tree * struct TreeNode {*     int val; *     TreeNode *left; *     TreeNode *right; *     Tre Enode (int x): Val (x), left (null), right (NULL) {} *}; */class Solution {public:    vector<int> inordertraversal (TreeNode *root) {        Stack<treenode *> Nodestack;        Vector<int> arr (0);        TreeNode *node;        if (root = NULL) return arr;        node = root;        while (!nodestack.empty () | | node!=null) {            if (node!=null) {                Nodestack.push (node);                node=node->left;            }            else{                node=nodestack.top ();                Arr.push_back (node->val);                Nodestack.pop ();                node=node->right;            }        }        return arr;    };

Binary Tree inorder Traversal