haskell - 用文件夹建立平衡的二叉树

我写了函数foldTree从列表中构建平衡的二叉树。
我必须使用foldr，没关系，我用过它，但是我使insertInTree函数递归=(现在我只知道这种穿过树的方式=))。

更新:我不确定函数insertTree:是否正确计算递归的高度？ =((这里需要一些帮助。

是否可以编写不递归的insertInTree(带有until/iterate/unfoldr的东西)，或者使foldTree函数不带有辅助函数=>更短？

这是我在下面的尝试:

data Tree a = Leaf
            | Node Integer (Tree a) a (Tree a)
            deriving (Show, Eq)

foldTree :: [a] -> Tree a
foldTree = foldr (\x tree -> insertInTree x tree) Leaf

insertInTree :: a -> Tree a -> Tree a
insertInTree x Leaf = Node 0 (Leaf) x (Leaf)
insertInTree x (Node n t1 val t2) = if h1 < h2
                                    then Node (h2+1) (insertInTree x t1) val t2
                                    else Node (h1+1) t1 val (insertInTree x t2)
  where h1 = heightTree t1
        h2 = heightTree t2

heightTree :: Tree a -> Integer
heightTree Leaf = 0
heightTree (Node n t1 val t2) = n

输出:

*Main> foldTree "ABCDEFGHIJ"
Node 3 (Node 2 (Node 0 Leaf 'B' Leaf) 'G' (Node 1 Leaf 'F' (Node 0 Leaf 'C' Leaf))) 'J' (Node 2 (Node 1 Leaf 'D' (Node 0 Leaf 'A' Leaf)) 'I' (Node 1 Leaf 'H' (Node 0 Leaf 'E' Leaf)))
*Main>

最佳答案

当两个子树的高度相等时，您的插入函数将出错，因为如果插入到正确的子树中，如果其已满，则会增加其高度。我现在还不清楚在您的代码中是否会出现这种情况。

在树中插入新元素的似乎正确的方法似乎是

insertInTree x (Node n t1 val t2)
    | h1 < h2   = Node  n (insertInTree x t1) val t2
    | h1 > h2   = Node  n    t1 val t2n
    | otherwise = Node (h+1) t1 val t2n
  where h1  = heightTree t1
        h2  = heightTree t2
        t2n = insertInTree x t2
        h   = heightTree t2n     -- might stay the same

这将创建几乎平衡的树(也称为AVL树)。但这会将每个新元素推到树的最底端。

编辑:这些树可以很好地看到

showTree Leaf = ""
showTree n@(Node i _ _ _) = go i n
  where
  go _ (Leaf) = ""
  go i (Node _ l c r) = go (i-1) l ++
    replicate (4*fromIntegral i) ' ' ++ show c ++ "\n" ++ go (i-1) r

尝试



是的，您可以将foldTree缩短为

foldTree = foldr insertInTree Leaf