lintcode:Binary Tree Level Order Traversal

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;    }};