Question

到目前为止，我已经想出了类似的东西。我在这里尝试做的是获取最右边或最左边的元素，具体取决于哪个可用，并用root交换它们并删除相应的最右边或最左边的元素。当我要求它删除树的根但是它适用于所有其他情况以及无可辩驳的模式失败意味着什么时，我只需要帮助找出它失败的原因？

如果我做delt 3 Node 3 (Node 2 (Node 1 Empty Empty) Empty) (Node 4 Empty Empty)之类的话它会出现Node *** Exception: delt.hs:26:40-75: Irrefutable pattern failed for pattern (Main.Node rm (Main.Empty) (Main.Empty))

之类的错误

delt 2 a提供Node 3 (Node 1 Empty Empty) (Node 4 Empty Empty)

data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show, Read, Eq)
treeIns :: (Ord a) => a -> Tree a -> Tree a
treeIns x Empty= Node x (Empty) (Empty)
treeIns x (Node a l r)
  | x == a = Node a l r
  | x < a = Node a (treeIns x l) r
  | x > a = Node a l (treeIns x r)

leftm :: Tree a -> Tree a
leftm Empty = Empty
leftm (Node a (Empty) (Empty)) = (Node a (Empty) (Empty))
leftm (Node a (l) (Empty)) = leftm l
leftm (Node a (l) (r)) = leftm l

rightm :: Tree a -> Tree a
rightm Empty = Empty
rightm (Node a (Empty) (Empty)) = (Node a (Empty) (Empty))
rightm (Node a (Empty) (r)) = rightm r
rightm (Node a (l) (r)) = rightm r


delt :: (Eq a, Ord a)=>a -> Tree a -> Tree a
delt x Empty = Empty
delt x (Node a (Empty)(Empty))
  | x== a = Empty
delt x (Node a l r)
  |x == a = (if l /= (Empty) then (let (Node rm (Empty) (Empty)) = rightm l in (Node rm (delt rm l) r)) else (let (Node rm (Empty) (Empty)) = l\
eftm r in (Node rm l (delt rm r)) ))
  |x>a = Node a (l) (delt x r)
  |x < a = Node a (delt x l ) r

Answer 1

这是一个类似的实现仅供参考。（虽然没有删除部分）

-- a tree can be empty or contain a value with two other Trees
data Tree a = EmptyTree | Node a (Tree a) (Tree a) deriving (Show, Read, Eq)

instance Functor Tree where
    fmap f EmptyTree = EmptyTree
    fmap f (Node x left right) = Node (f x) (fmap f left) (fmap f right)

-- create a Node with a value and two Empty subtrees (left and right)
singleton :: a -> Tree a
singleton x = Node x EmptyTree EmptyTree

-- insert a new Node into a tree
insertInTree :: (Ord a) => a -> Tree a -> Tree a

-- insert into empty tree is equal to creating a new tree
insertInTree x EmptyTree = singleton x

-- pattern match on value and subtrees
insertInTree x (Node a left right)
    -- if element is equal to root element, return same tree
    | x == a = Node x left right

    -- if element is smaller than root, go to left
    -- subtree and check again.
    | x < a = Node a (insertInTree x left) right

    -- if element is bigger than root, go to right
    -- subtree and check again.
    | x > a = Node a left (insertInTree x right)

-- binary tree search
search :: (Ord a) => a -> Tree a -> Bool

-- search in EmptyTree is always False
search x EmptyTree = False

-- serach in a Node
search x (Node a left right)

    -- if element equals root element, great
    | x == a = True

    -- if element is smaller than root, continue search on the left side
    | x < a = search x left

    -- if element is bigger than root, continue search on the right side
    | x > a = search x right

-- create a test tree
myTree = foldr insertInTree EmptyTree [15,75,651,2,3,4,85,42,1,5,36,45,78,12,2]

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