Binary Tree Inorder Traversal [LEETCODE]

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? 

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}".

 

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Easy bin-tree traverse problem. 

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> result;
        if(NULL == root){
            return result;
        }
        
        const int BOTH_NOT_VISITED = 0;
        const int ONLY_LEFT_VISIED = 1;
        const int ONLY_RIGHT_VISITED = 2;
        const int BOTH_VISITED = 3;
        stack<pair<TreeNode *, int> > s;
        s.push(make_pair(root, BOTH_NOT_VISITED));
        while(!s.empty()) {
            pair<TreeNode *, int> *p = &(s.top());
            if(0 == p->second) {
                //visit left
                if(NULL != p->first->left){
                    s.push(make_pair(p->first->left, BOTH_NOT_VISITED));
                }
                p->second = ONLY_LEFT_VISIED;
            } else if(ONLY_LEFT_VISIED == p->second) {
                //visit middle
                result.push_back(p->first->val);
                p->second = ONLY_RIGHT_VISITED;
            } else if(ONLY_RIGHT_VISITED == p->second){
                //visit right
                if(NULL != p->first->right){
                    s.push(make_pair(p->first->right, BOTH_NOT_VISITED));
                }
                p->second = BOTH_VISITED;
            }
            else{
                s.pop();
            }
        }
        return result;
    }
};

Remember to use pointer when get the top element of stack.