[leetcode] 9. Binary Tree Level Order Traversal

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);

		}

	}

};