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? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

 /**  
  * Definition for binary tree  
  * struct TreeNode {  
  *   int val;  
  *   TreeNode *left;  
  *   TreeNode *right;  
  *   TreeNode(int x) : val(x), left(NULL), right(NULL) {}  
  * };  
  */ 

 
   void inorder(TreeNode *root,int level, vector<vector<int> > &v)  
   {  
     if(!root)return;  
     inorder(root->left,level+1,v);  
     v[level].push_back(root->val);  
     inorder(root->right,level+1,v);  
   }  

   int depth(TreeNode *root)  
   {  
     if(!root)return 0;  
     return (max(depth(root->left),depth(root->right))+1);  
   }  

   vector<vector<int> > levelOrder(TreeNode *root) {  
     vector<vector<int> > v(depth(root));  
     inorder(root,0, v);  
     return v;  
   }