Convert Linked List to Binary Search Tree

Problem: Convert Linked List to Binary Search Tree

The general idea is to find the center node of linked list and make it root of a tree. Then we do the same thing to its left part and right part to get left and right sub-tree. To find center node of a linked list, we can use two pointers, one going 2 steps ahead and one going 1 step. When the fast pointer gets to the end, the slow pointer gets to the middle node.

Code in Python:

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        if not head: return None
        def helper(head):
            if not head: return None
            if not head.next:
                root = TreeNode(head.val)
                return root
            slow = fast = head
            prev = None
            while fast and fast.next:
                prev = slow
                slow = slow.next
                fast = fast.next.next
            prev.next = None
            root = TreeNode(slow.val)
            root.left, root.right = helper(head), helper(slow.next)
            return root
        return helper(head)