# 单向链表转平衡二叉搜索树

#### 题目描述

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

##### 我的笨代码
/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* sortedListToBST(ListNode* head) {
return NULL;

vector<int> temp;
{
}

return vecToBST(temp);

}

private:
TreeNode* vecToBST(vector<int> &node)
{
if(node.size() == 0)
return NULL;

int len = node.size();
TreeNode *root = new TreeNode(node[ len/2 ]);
vector<int> left(node.begin(), node.begin() + len/2);
vector<int> right(node.begin() + len/2 + 1, node.end());
root->left = vecToBST(left);
root->right = vecToBST(right);

return root;
}
};


##### 参考别人的代码

（平衡）二叉树搜索树的中序变量可以是一个排了序的单向链表， 可以用左右子树的节点个数，控制二叉搜索树是一个平衡二叉搜索树。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
ListNode* list;

public:
TreeNode* sortedListToBST(ListNode* head) {
return NULL;

}

private:
// 计数
int count(ListNode* pos)
{
int size = 0;
while(pos != NULL)
{
size++;
pos = pos->next;
}

return size;
}

// 按照计数和中序遍历构造平衡二叉搜索树
TreeNode* generate(int n)
{
if(n == 0)
return NULL;

TreeNode* node = new TreeNode(0);
node->left = generate(n/2);
node->val = list->val;
list = list->next;
node->right = generate(n - n/2 - 1);

return node;
}
};



posted @ 2017-07-04 23:00  草滩小恪  阅读(72)  评论(0编辑  收藏