小编use*_*968的帖子

假设一个二叉搜索树,我想在我们尝试插入已经存在的元素时返回错误.有没有办法让这项工作？

data BST2 a = EmptyBST2 | Node2 a (BST2 a) (BST2 a)  deriving Show

insert2 :: a -> Either b (BST2 a) -> Either b (BST2 a)
insert2 elem (Right EmptyBST2) = Right (Node2 elem EmptyBST2 EmptyBST2)
insert2 elem (Right (Node2 root left right))
  | (elem == root) = Left "Error: Element already exist."
  | (elem < root) = (Node2 root (insert2 elem left) right)
  | otherwise = (Node2 root left (insert2 elem right))

注意:我是Haskell的新手.