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[leetcode] 7. Binary Tree Level Order Traversal II

时间:2014-11-15 01:27:53      阅读:174      评论:0      收藏:0      [点我收藏+]

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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word发布,测试一下。

[leetcode] 7. Binary Tree Level Order Traversal II

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原文地址:http://www.cnblogs.com/TinyBox/p/4098580.html

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