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]
]

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<vector<int> > levelOrder(TreeNode *root) 
    {
        vector<vector<int>> res;
        if(root == NULL)
            return res;
        int save;
        queue<TreeNode*> use;
        use.push(root);
        save = 1;
        while(save!=0)
        {
            int count = save;
            save = 0;
            vector<int> t;
            while(count > 0 && !use.empty())
            {
                TreeNode* temp = use.front();
                use.pop();
                count--;
                t.push_back(temp->val);
                if(temp->left!=NULL)
                {
                    use.push(temp->left);
                    save++;
                }
                if(temp->right!=NULL)
                {
                    use.push(temp->right);
                    save++;
                }
            }
            res.push_back(t);
        }
        return res;
    }
};

LeetCode--Binary Tree Level Order Traversal