public static TreeNode sortedListToBST(ListNode head)
{
if (head == null)
return null;
ListNode cur = head;
int count = 0;
while (cur != null)
{
cur = cur.next;
count++;
}
return helper(head, 0, count - 1);
}
private static TreeNode helper(ListNode list, int l, int r)
{
if (l > r )
return null;
int m = (l + r) / 2;
TreeNode left = helper(list, l, m - 1);
TreeNode root = new TreeNode(list.val);
root.left = left;
list = list.next;
root.right = helper(list, m + 1, r);
return root;
}
private static TreeNode helper(ListNode list, int l, int r)
{
if (l > r)
{
return(null);
}
int m = (l + r) / 2;
TreeNode left = helper(list, l, m - 1);
TreeNode root = new TreeNode(list.val);
root.left = left;
list = list.next;
root.right = helper(list, m + 1, r);
return(root);
}
/* study the sample code: http://codeganker.blogspot.ca/2014/04/convert-sorted-list-to-binary-search.html
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
*/
static void Main(string[] args)
{
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
// head.next.next.next = new ListNode(4);
// head.next.next.next.next = new ListNode(5);
// head.next.next.next.next.next = new ListNode(6);
// head.next.next.next.next.next.next = new ListNode(7);
TreeNode root = sortedListToBST(head);
}