我一直在嘗試解決有關二叉搜索樹的遞歸問題，但是我沒有運氣。有人可以用最簡單的形式向我解釋一下這段代碼（在這個問題中廣泛使用）如何將數組轉換為 BST：def helper(left, right):            # base case            if left > right:                return None整個代碼（取自leetcode https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/discuss/900790/Python3-with-explanation-faster-than-100-PreOrder-traversal）def sortedArrayToBST(self, nums: List[int]) -> TreeNode:        # statrt with the middle element and for the right ones go tree.right and for the left ones go tree.left        # would have to traverse once so the time complexity will be O(n).        def helper(left, right):            # base case            if left > right:                return None                        # get the length to the nearest whole number            length  = (left + right) // 2                        # preOrder traversal            root = TreeNode(nums[length])            root.left = helper(left , length -1)            root.right = helper(length+1, right)            return root                return helper(0 , len(nums) - 1)對此事的任何幫助都會很棒！謝謝