获取玫瑰树中节点的父节点

Question

我正在尝试使用以下代码获取玫瑰树中节点的父节点：

{-# LANGUAGE DeriveAnyClass    #-}
{-# LANGUAGE DeriveFunctor     #-}
{-# LANGUAGE DeriveGeneric     #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE OverloadedStrings #-}

module Main where

import GHC.Generics (Generic)

data RoseTree a = RoseTree { value :: a, children :: [RoseTree a]}
                   deriving (Show, Generic, Functor)

type Timestamp = Integer

type Tree = RoseTree Timestamp 

getParentNode :: Timestamp -> Tree -> Maybe Timestamp 
getParentNode node tree@(RoseTree _ []) = Nothing
getParentNode node tree@(RoseTree rootNode (x:xs)) =
  if node == value x then Just rootNode else case getParentNode node x of 
    Nothing -> case getParentNodeForList node (children x) of 
      Nothing -> getParentNodeForList node xs
      Just parent -> Just parent
    Just parent -> Just parent 

getParentNodeForList :: Timestamp -> [Tree]-> Maybe Timestamp
getParentNodeForList node [] = Nothing
getParentNodeForList node (x:xs) = case getParentNode node x of 
  Nothing -> case getParentNodeForList node (children x) of 
    Nothing -> getParentNodeForList node xs
    Just parent -> Just parent
  Just parent -> Just parent 


main :: IO ()
main = do
  let tree = RoseTree 1623839394 [RoseTree 1623839395 [], RoseTree 1623839396 [], RoseTree 1623839397 []]
  putStrLn $ show $ getParentNode 1623839397 tree

对于这种情况， Nothing是空的。 我不懂为什么。 我在getParentNode和getParentNodeForList中涵盖了所有可能的遍历情况（至少看起来是这样，它显然不起作用）。

任何帮助表示赞赏。

Answer 1

我认为问题是getParentNode应该在getParentNode node (RoseTree rootNode xs)上递归，所以整个 function 会变成：

getParentNode :: Timestamp -> Tree -> Maybe Timestamp 
getParentNode node tree@(RoseTree _ []) = Nothing
getParentNode node tree@(RoseTree rootNode (x:xs)) =
  if node == value x then Just rootNode else case getParentNode node x of 
    Nothing -> case getParentNodeForList node (children x) of 
      Nothing -> getParentNode node (RoseTree rootNode xs)
      Just parent -> Just parent
    Just parent -> Just parent 

就个人而言，我会使用asum来更简洁地编写它：

getParentNode node (RoseTree x xs)
  | node `elem` map value xs = pure x
  | otherwise = asum (map (getParentNode node) xs)