Algorithm 具有键值的Haskell二叉树
我想用具有元组(k,v)的键值叶构建二叉树 我的代码:Algorithm 具有键值的Haskell二叉树,algorithm,haskell,binary-search-tree,Algorithm,Haskell,Binary Search Tree,我想用具有元组(k,v)的键值叶构建二叉树 我的代码: data Tree k v = EmptyTree | Node (k, v) (Tree k v) (Tree k v) deriving (Show, Eq, Ord, Read) emptyTree :: (k,v) -> Tree k v emptyTree (k,v) = Node (k, v) EmptyTree EmptyTree treeInse
data Tree k v = EmptyTree
| Node (k, v) (Tree k v) (Tree k v)
deriving (Show, Eq, Ord, Read)
emptyTree :: (k,v) -> Tree k v
emptyTree (k,v) = Node (k, v) EmptyTree EmptyTree
treeInsert :: (Ord k) => (k,v) -> Tree k v -> Tree k v
treeInsert (k,v) EmptyTree = emptyTree (k, v)
treeInsert (a, b) (Node (k,v) left right)
| a == k = (Node (a,b) left right)
| a < k = (Node (a, b) (treeInsert (a, b) left) right)
| a > k = (Node (a, b) left (treeInsert (a, b) right))
但我得到了这个错误:
Couldn't match type `v' with `Int'
`v' is a rigid type variable bound by
the type signature for fillTree :: Int -> Tree k v -> Tree k v
原因是什么?我如何修复它?您的类型要么太笼统,要么太具体。应该是
fillTree :: Int -> Tree Int Int -> Tree Int Int
或
您的原始声明试图为任何k
,v
将(Int,Int)
插入树kv
。它是说无论你有什么样的树,我们都可以在其中插入一对Int
s。这显然是胡说八道,正如您对treeInsert
的签名所示,只有成对的(k,v)
类型才能插入到树k v
中
treeInsert :: (Ord k) => (k, v) -> Tree k v -> Tree k v
或者,类似于
(Ord a)=>a->Tree a a->Tree a a
。在这种情况下,删除类型签名、在GHCi中加载文件并查看编译器认为类型应该是什么可能会有所帮助。emptyTree
对于该函数来说是一个非常糟糕的名称,每个人都会期望它返回一个EmptyTree
。更好的名称应该是singleton
或singleNode
。
fillTree :: (Ord a) => a -> Tree a a -> Tree a a
treeInsert :: (Ord k) => (k, v) -> Tree k v -> Tree k v