# Question

Given 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:

```
Given the sorted linked list: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
```

# Solution

Use recursion to maintain tree order, but pass the next node in the list around so that the time complexity is \(O(N)\).

```
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def sortedListToBST(self, head):
"""
:type head: ListNode
:rtype: TreeNode
"""
if head is None:
return None
iterator = head
node = head
size = 0
while iterator is not None:
iterator = iterator.next
size += 1
def builder(start, end, node):
if end < start:
return (None, node)
elif end == start:
root = TreeNode(node.val)
return (root, node.next)
else:
mid = (start + end) // 2
left, new_node = builder(start, mid - 1, node)
root = TreeNode(new_node.val)
new_node = new_node.next
right, new_node = builder(mid + 1, end, new_node)
root.left = left
root.right = right
return (root, new_node)
return builder(0, size - 1, node)[0]
```