我来自Java背景,想学习一些Haskell。但是,此刻有点停滞不前。
我想要做的是:我有一个树列表,其中每个节点在列表中的所有树中都有唯一的标识符。现在我想从其中一个树中删除一个节点并返回新树以及未更改的树。
删除节点应该:
想象一下以下树木:
当我删除节点' 2'我希望结果是以下树:
树中的每个节点都包含标识符和子树列表。 这是我到目前为止所做的,但它显然不起作用,我对如何用Haskell解决这个问题感到有点迷失:
import Data.Tree
data CustomNode = CustomNode { identifier :: Int } deriving (Ord,Eq,Show,Read)
type CustomTree = Tree CustomNode
myTree0 = t0
where
leaf i = Node CustomNode{identifier = i} []
t0 = Node CustomNode{identifier = 0} [t1]
t1 = Node CustomNode{identifier = 1} [t2, t5]
t2 = Node CustomNode{identifier = 2} [leaf 3, leaf 4]
t5 = Node CustomNode{identifier = 5} [leaf 6]
myTree1 = t0
where
leaf i = Node CustomNode{identifier = i} []
t0 = Node CustomNode{identifier = 7} [leaf 8]
deleteNode :: Int -> [CustomTree] -> [CustomTree]
deleteNode _ [] = []
deleteNode n (x:xs) = if isNodeInTree n x then deleteNodeFromTree n x ++ xs else deleteNode n xs
--below is the fixed line as per the answer below
--deleteNode n (x:xs) = if isNodeInTree n x then deleteNodeFromTree n x ++ xs else x : deleteNode n xs
deleteNodeFromTree :: Int -> CustomTree -> [CustomTree]
deleteNodeFromTree n (Node c xs) = if identifier c == n then [] else deleteNode n xs
--below is the fixed line as per the answer below
--deleteNodeFromTree n (Node c xs) = if identifier c == n then xs else deleteNode n xs
isNodeInTree :: Int -> CustomTree -> Bool
isNodeInTree n (Node c xs) = if identifier c == n then True else isNodeInForest n xs
isNodeInForest :: Int -> [CustomTree] -> Bool
isNodeInForest n [] = False
isNodeInForest n (x:xs) = if isNodeInTree n x then True else isNodeInForest n xs
非常感谢任何帮助!
答案 0 :(得分:5)
看起来你在这里已经有了一个合理的开始。
我假设您希望deleteNode
获取林并返回相同的林,但删除了指定的节点。在那种情况下,
if isNodeInTree n x then ... else deleteNode n xs
你把x
扔掉了。你可能并不是故意这样做。
... else x : deleteNode n xs
会把那棵树留在森林里,这可能就是你想要的了。
此外,在deleteNodeFromTree
:
if identifier c == n then [] ...
也许你想在此时返回节点的所有子节点? (所以他们都成了根节点。)
这些是我唯一能够脱颖而出的东西。看看你需要的地方......