[leetcode] 7. Binary Tree Level Order Traversal II

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。