108.将有序数组转换为平衡二叉树

思路：递归+二分每个节点作为根节点的时候，左子树形成的数组一定比它小，右子树形成的数组一定比它大，符合有序数组中点的性质

class Solution(object):
    def sortedArrayToBST(self, nums):
        begin = 0
        end = len(nums)-1
        if begin>end:
            return None
        mid = (begin+end)//2
        root = TreeNode(nums[mid])
        root.left = Solution().sortedArrayToBST(nums[begin:mid])
        root.right = Solution().sortedArrayToBST(nums[mid+1:end+1])
        return root