helper(left, right) that takes the indices of the current subarray as arguments.left > right, the subarray is empty, and we return None.p = (left + right) // 2.root with the value nums[p].left to p - 1.p + 1 to right.root.left and root.right respectively.root node.helper(0, len(nums) - 1) to initiate the recursion and construct the BST.