Convert Sorted Array to Binary Search Tree - （递归建树）

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

我写的代码。。。。。。。。。。

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        n = len(nums)
        mid = n/2
        low = 0
        high = n-1
        return self.buildBST(low, mid, high, nums)
        
    def buildBST(self, low, mid, high, nums):
        if low > mid or mid > high:
            return None
        node = TreeNode(nums[mid])
        node.left = self.buildBST(low, int((mid-low)/2)+low, mid-1, nums)
        node.right = self.buildBST(mid+1, int((high-mid)/2)+mid+1, high, nums)
        return node

别人写的代码。。。。。。。。。。。ε=(´ο｀*)))唉

class Solution(object):
    def sortedArrayToBST(self, nums):
        if not nums:
            return None
        n=len(nums)
        root = TreeNode(nums[n/2])
        root.left = self.sortedArrayToBST(nums[:n/2])
        root.right=self.sortedArrayToBST(nums[n/2+1:])
        return root