把排序数组转换为高度最小的二叉搜索树

LeetCode题目地址
思路解释

def toBST(self, A, start, end):
        if start < 0 or end < 0 or start > end:
            return None
        mid = (start + end) / 2
        root = TreeNode(A[mid])
        #if start != end:
        root.left = self.toBST(A,start,mid-1)
        root.right = self.toBST(A,mid+1,end)
        return root
    def sortedArrayToBST(self, A):
        # write your code here
        return self.toBST(A,0,len(A)-1)