Haskell 从树中删除节点并返回结果林
我来自Java背景,想学习一些Haskell。不过,现在有点卡住了 我想做的是:我有一个树列表,其中每个节点在列表中的所有树上都有一个唯一的标识符。现在我想从这些树中删除一个节点,并返回新树和未更改的树 删除节点应:Haskell 从树中删除节点并返回结果林,haskell,tree,Haskell,Tree,我来自Java背景,想学习一些Haskell。不过,现在有点卡住了 我想做的是:我有一个树列表,其中每个节点在列表中的所有树上都有一个唯一的标识符。现在我想从这些树中删除一个节点,并返回新树和未更改的树 删除节点应: 使所述节点的所有子节点成为新树的根 删除已删除节点(包括根节点)的父节点,并对所有已删除节点执行上述操作 想象一下下面的树: 删除节点“2”时,我希望结果为以下树: 树中的每个节点都由标识符和子树列表组成。 到目前为止,我所掌握的就是这些,但它显然不起作用,我对如何用Has
- 使所述节点的所有子节点成为新树的根
- 删除已删除节点(包括根节点)的父节点,并对所有已删除节点执行上述操作
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
任何帮助都将不胜感激 看来你在这里已经有了一个合理的开始 我假设您希望
deleteNode
获取一个林并返回同一个林,但删除指定的节点。那么,
if isNodeInTree n x then ... else deleteNode n xs
你刚刚扔掉了x
。你可能不是故意的
... else x : deleteNode n xs
我会把那棵树留在森林里,这可能是你想要的
此外,在deleteNodeFromTree
中:
if identifier c == n then [] ...
也许您想在此时返回节点的所有子节点?(因此它们都成为根节点。)
这些是我唯一能突出的东西。看看这会带你去哪里…@MathematicalOrchid是的。对不起,我忘了添加导入。令人惊讶的是,这两件东西是我所缺少的所有东西!我想这会是更多的工作!非常感谢,哈斯克尔实际上非常棒:)@Marco Happy Haskling.。-)