sortedListToBST.py

#    -*-    coding: utf-8    -*-
'''
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
'''
"""
        :type head: ListNode
        :rtype: TreeNode
        """
# 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 sortedArrayToBST(self, nums):
        if len(nums) == 0:
            return None
        elif len(nums) == 1:
            return TreeNode(nums[0])
        mid = len(nums)//2
        root = TreeNode(nums[mid])
        left = self.sortedArrayToBST(nums[:mid])
        right = self.sortedArrayToBST(nums[mid+1:])
        root.left = left
        root.right = right
        return root
    def sortedListToBST(self, head):
        nums = []
        while head != None:
            nums.append(head.val)
            head = head.next
        return self.sortedArrayToBST(nums)

if __name__ == "__main__":
	sol = Solution()