跟第七题一样，把最后的输出顺序换一下就行。。。

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.

题解如下：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:	vector<vector<int> > levelOrder(TreeNode *root) 	{		vector<vector<int>> temp;		int len = MaxDepth(root);						for (int  i = 0; i < len; i++)		{			vector<int> level;			getElement(level, 0, i, root);			temp.push_back(level);			level.clear();		}		return temp;	}	int MaxDepth(TreeNode *temp)	{	 	if (temp == NULL)	 		return 0;	 	else	 	{	 		int aspros = MaxDepth(temp->left);	 		int defteros = MaxDepth(temp->right);	 		return 1 + (aspros>defteros ? aspros : defteros);	 	}	}	void getElement(vector<int> &level, int count, int len, TreeNode *root)	{		if (root != NULL)		{			if (count == len)			{				level.push_back(root->val);			}			getElement(level, count + 1, len, root->left);			getElement(level, count + 1, len, root->right);		}	}};

[leetcode] 9. Binary Tree Level Order Traversal