Convert Sorted List to Binary Search Tree

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example 1:

Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.

Example 2:

Input: head = []
Output: []

Constraints:

• The number of nodes in head is in the range [0, 2 * 104].
• -105 <= Node.val <= 105

SOLUTION:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedListToBST(self, head: Optional[ListNode]) -> Optional[TreeNode]:
        if not head:
            return head
        if not head.next:
            return TreeNode(val = head.val)
        prev = None
        slow = head
        fast = head
        while slow and fast and fast.next:
            prev = slow
            slow = slow.next
            fast = fast.next.next
        prev.next = None
        root = TreeNode(val = slow.val)
        root.left = self.sortedListToBST(head)
        root.right = self.sortedListToBST(slow.next)
        return root