这次相对来讲复杂点，题目如下：

Given a binary tree, return the bottom-up level order traversal of its nodes‘ values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   /   9  20    /     15   7

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]]

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}".

就是按照层数来输出树。思路分为三步，先拿到树的最大层数len，然后再写一个拿指定层的各元素的函数，最后按题目要求写一个返回vector<vector<int>>的函数，里面用一个循环来多次调用之前那个函数，好了，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> > levelOrderBottom(TreeNode *root) 	{		vector<vector<int>> temp;		int len = MaxDepth(root);						for (int  i = len - 1; i >= 0; 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);		}	}};

因为题目要求是要从底部向上输出，所以在主函数的for循环里用了倒序。

这次的blog用Live Writer发布，测试一下。

[leetcode] 7. Binary Tree Level Order Traversal II