题目:
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
思路:
类似
http://blog.csdn.net/lanxu_yy/article/details/11898523,不过需要利用链表的方式找到中间节点。
代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
ListNode* getMid(ListNode *head, ListNode *tail)
{
if(head == tail)
return NULL;
else
{
ListNode * slow=head;
ListNode * fast=head;
while(fast != tail && fast->next != tail)
{
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
}
TreeNode *sortedListToBST(ListNode *head) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return sortedArrayToBST(head, NULL);
}
TreeNode *sortedArrayToBST(ListNode *head, ListNode *tail)
{
ListNode * p = getMid(head, tail);
if(p == NULL)
{
return NULL;
}
else
{
TreeNode * parent = new TreeNode(p->val);
parent->left = sortedArrayToBST(head, p);
parent->right = sortedArrayToBST(p->next, tail);
return parent;
}
}
};