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},

 1 struct TreeNode { 2     int        val; 3     TreeNode*    left; 4     TreeNode*    right; 5     TreeNode():val(0),left(NULL),right(NULL) {} 6     TreeNode(int x): val(x), left(NULL),right(NULL) {} 7 }; 8  9 10 vector<vector<int> > levelOrderTraversal(TreeNode *root) //非递归的中序遍历(用栈实现)11 {12 13     queue<TreeNode *> tree_queue;14     vector<vector<int> > tree_vector;15     vector<int> svector;16 17     if (NULL == root) {18         return tree_vector;19     }20     TreeNode *pTemp = root;21     tree_queue.push(root);22     tree_queue.push(NULL); //the end of one level.23 24     while (true) {25         TreeNode *pTemp = tree_queue.front();26         tree_queue.pop();27 28         if (!pTemp) { //get the null, put vector<> to vector<vector<>>29             tree_vector.push_back(svector);30             svector.clear();31             if (tree_queue.empty())32                 break;33             tree_queue.push(NULL);34         }35         else {36             svector.push_back(pTemp->val);37             if (pTemp->left) 38                 tree_queue.push(pTemp->left);39             if (pTemp->right)40                 tree_queue.push(pTemp->right);41         }42     }43     return tree_vector;44 }