Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
/**
* 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) {}
* };
*/
struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) :val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
TreeNode *sortedListToBST(ListNode *head) {
int n = 0;
ListNode *p = head;
while(p != NULL)
{
p = p->next;
n++;
}
return ListToBST(head,0,n-1);
}
private:
TreeNode *ListToBST(ListNode *&head,int start, int end)
{
if(start > end) return NULL;
int mid = (start + end) / 2;
TreeNode *left = ListToBST(head,start,mid - 1);
TreeNode *node = new TreeNode(head->val);
node->left = left;
head = head->next;
node->right = ListToBST(head,mid + 1,end);
return node;
}
};
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