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109. Convert Sorted List to Binary Search Tree

Difficulty Topics

Description

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

convert-sorted-list-to-binary-search-tree.py
# 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: ListNode) -> TreeNode:
        if not head:
            return None 
        if not head.next:
            return TreeNode(head.val)

        pre, slow, fast = None, head, head

        # when fast is at the end, slow will be at mid 
        while fast and fast.next:
            pre, slow, fast = slow, slow.next, fast.next.next
        if pre:
            # cut the parts from mid 
            pre.next = None


        root = TreeNode(slow.val)
        root.left = self.sortedListToBST(head)
        root.right = self.sortedListToBST(slow.next)

        return root