Binary Tree Inorder Traversal

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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.

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(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