[leetcode] 9. Binary Tree Level Order Traversal

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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

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原文地址：http://www.cnblogs.com/TinyBox/p/4100205.html