leetcode - 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).

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?

struct TreeNode{int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    std::vector<std::vector<int> > levelOrder(TreeNode *root){        //第一种解法，利用队列BFS分层遍历二叉树std::vector<std::vector<int>> vec;        if(root == NULL) return vec;std::queue<TreeNode*> Q;Q.push(root);int cnt = 1;std::vector<int> tmp;while(!Q.empty()){tmp.clear();int level = 0;for(int i = 0; i < cnt; i++){root = Q.front();Q.pop();tmp.push_back(root->val);if(root->left != NULL) Q.push(root->left),level++;if(root->right != NULL) Q.push(root->right),level++;}cnt = level;vec.push_back(tmp);}return vec;    }};

class Solution {public:    std::vector<std::vector<int> > levelOrder(TreeNode *root){        //第二种方法也是利用队列的方法来做的，但是，其实数据结构的实现是vector,觉得这种方法更简便一些.std::vector<std::vector<int>> result;if(root == NULL) return result;std::vector<int> vec;std::vector<TreeNode*> node;node.push_back(root);int cur = 0,last = 1;while(cur < node.size()){last = node.size();while(cur < last){vec.push_back(node[cur]->val);if(node[cur]->left != NULL) node.push_back(node[cur]->left);if(node[cur]->right != NULL) node.push_back(node[cur]->right);cur++;}result.push_back(vec);vec.clear();}return result;    }};

class Solution {public:    std::vector<std::vector<int> > levelOrder(TreeNode *root){        //第三种解法，dfs分层遍历二叉树dfs(root,0);return result;    }private:std::vector<std::vector<int>> result;void dfs(TreeNode *root,int level){if(root == NULL) return;if(level == result.size()){std::vector<int> vec;result.push_back(vec);}result[level].push_back(root->val);dfs(root->left,level+1);dfs(root->right,level+1);}};