LeetCode #109 Convert Sorted List to Binary Search Tree

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Problem

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

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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

Solution1

Method: Divide and Conquer

Time Complexity: O(n log n)

Space Complexity: O(n)

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# 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 sortedListToArray(self, head):
        """
        :type head: ListNode
        :rtype: array
        """
        ret = []
        while head:
            ret.append(head.val)
            head = head.next

        return ret

    def sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        if not head:
            return []

        array = self.sortedListToArray(head)
        tree_head = self.helper(array)

        return tree_head

    def helper(self, array):
        """
        :type array: array
        :rtype: TreeNode
        """
        if not array:
            return None
        elif len(array) == 1:
            return TreeNode(array[0])
        elif len(array) == 2:
            tree_node = TreeNode(array[0])
            tree_node.right = TreeNode(array[1])
            return tree_node

        mid = len(array) //2
        tree_node = TreeNode(array[mid])
        tree_node.left = self.helper(array[:mid])
        if mid + 1 < len(array):
            tree_node.right = self.helper(array[mid + 1:])

        return tree_node

or

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# 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):
        if not head:
            return 
        if not head.next:
            return TreeNode(head.val)
            
        # here we get the middle point,
        # even case, like '1234', slow points to '2',
        # '3' is root, '12' belongs to left, '4' is right
        # odd case, like '12345', slow points to '2', '12'
        # belongs to left, '3' is root, '45' belongs to right
        slow, fast = head, head.next.next
        
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
        
        # tmp points to root
        tmp = slow.next
        
        # cut down the left child
        slow.next = None
        root = TreeNode(tmp.val)
        root.left = self.sortedListToBST(head)
        root.right = self.sortedListToBST(tmp.next)
        
        return root

Reference

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/discuss/35474/Python-recursive-solution-with-detailed-comments-(operate-linked-list-directly).