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

Challenge 1: Using only 1 queue to implement it.

Challenge 2: Use DFS algorithm to do it.

1.队列
以前一直依次输出每层的节点标号，很简单。但现在需要把每层的标号存在一个二维向量中。
可以在每一层末尾插入一个NULL来区分层次！

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */


class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Level order a list of lists of integer
     */
public:
    vector<vector<int>> levelOrder(TreeNode *root) {
        // write your code here
        vector<vector<int> > res;

        if(root==NULL){
            return res;
        }

        queue<TreeNode*> Q;
        Q.push(root);
        Q.push(NULL);

        vector<int> level;

        while(!Q.empty()){
            TreeNode *node=Q.front();
            Q.pop();

            if(node){
               level.push_back(node->val);

               if(node->left){
                  Q.push(node->left);
               }

               if(node->right){
                 Q.push(node->right);
               }
            }else{

                res.push_back(level);
                level.clear();

                if(!Q.empty()){
                    Q.push(NULL);
                }
            }

        }

        return res;
    }
};