Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3

   / \

  9  20

    /  \

   15   7

return its level order traversal as:

[

  [3],

  [9,20],

  [15,7]

]

vector<vector<int> > levelOrder(TreeNode *root) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

        vector<vector<int>> matrix;

        if(root == NULL) return matrix;

		vector<int> r;

		r.push_back(root->val);

		matrix.push_back(r);

		vector<TreeNode *> path;

		path.push_back(root);

		int count = 1;

		while(!path.empty())

		{

			TreeNode* tn = path.front();

			if(tn->left) path.push_back(tn->left);

			if(tn->right) path.push_back(tn->right);

			path.erase(path.begin());

			count--;

			if(count==0)

			{

				vector<int> tmp;

				vector<TreeNode*>::iterator it = path.begin();

				for(;it != path.end(); ++it)

					tmp.push_back((*it)->val);

                if(tmp.size() > 0)

				    matrix.push_back(tmp);

				count = path.size();

			}

		}

		return matrix;

    }