108ConvertSortedArrayToBinarySearchTree.py

"""
Given an array 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 array: [-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
"""
"""
Comments
"""
"""
My
"""
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
        n = len(nums)
        if n  == 0:
            return None
        r = n // 2
        root = TreeNode(nums[r])
        
        def getNode(begin, end):
            if not begin<=end:
                return None
            i = (begin+end) // 2
            node = TreeNode(nums[i])
            node.left = getNode(begin, i-1)
            node.right = getNode(i+1, end)
            return node
        root.left = getNode(0,r-1)
        root.right = getNode(r+1, n-1)
        return root
"""
Fast
"""
class Solution:
    def sortedArrayToBST(self, nums: 'List[int]') -> 'TreeNode':
        if not nums:
            return None
        
        return self.toBstUtil(nums, 0, len(nums) - 1)
    
    def toBstUtil(self, nums, low, high):
        
        if low > high:
            return None
        
        mid = (low + high) // 2
        
        node = TreeNode(nums[mid])       
        node.left = self.toBstUtil(nums, low, mid - 1)
        node.right = self.toBstUtil(nums, mid + 1, high)
        
        return node