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我正在尝试使用 F# 对简单的树结构进行建模，并且不禁认为我做得非常糟糕：

我的树本质上是一个叶子列表（最终将被持久化到数据库表中）。我有一个函数 getChildren 接收叶节点 ID 并递归返回该叶的所有子节点。

open System.Collections.Generic

type leaf = { nodeID : int; nodeDescr : string; parentID : int option}

let myTree = [ { nodeID = 0;  nodeDescr = "Root"; parentID = None };
                 { nodeID = 1;  nodeDescr = "Mechanical"; parentID = Some(0) } ;
                 { nodeID = 2;  nodeDescr = "Electrical"; parentID = Some(0) } ;
                 { nodeID = 3;  nodeDescr = "High Voltage"; parentID = Some(2) } ;
                 { nodeID = 4;  nodeDescr = "Low Voltage"; parentID = Some(2) } ;
                 { nodeID = 5;  nodeDescr = "HV Maintanence"; parentID = Some(3) } ;
                 { nodeID = 6;  nodeDescr = "City Power"; parentID = Some(3) } ;
                 { nodeID = 7;  nodeDescr = "LV Wiring"; parentID = Some(4) } ;
                 { nodeID = 8;  nodeDescr = "LV Maintanence"; parentID = Some(4) } ]


let getChildren (id : int) (tree : list<leaf>) = 
    let allChildren = new List<leaf>() // Mutable list

    let rec getAllChildren (id : int) (t : list<leaf>) = 
        let cl = List.filter (fun x -> x.parentID = Some id) t // Get the immediate children
        for c in cl do // Loop through the immediate children and recursively get their children
            allChildren.Add(c)
            getAllChildren c.nodeID t
    getAllChildren id tree
    allChildren

我在这里的担忧是：

  1. 我正在使用可变列表
  2. 我正在使用循环

我怀疑在 F# 中使用函数式编程时有一种更优雅的方法可以避免突变和循环，而且我的命令式编程习惯正在潜移默化。

此外，这是对树结构建模的好方法吗，请记住它需要从数据库表中存储和检索？

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1 回答 1

4

如果你想保留你已经拥有的树结构，这个函数会为你找到孩子，没有循环或可变值：

let getChildren (id : int) (tree : list<leaf>) = 
    let parent node = tree |> Seq.filter (fun x -> Some x.nodeID = node.parentID) |> Seq.exactlyOne

    let rec hasAncestor (node : leaf) =
        node.parentID = Some id || (node.parentID.IsSome && hasAncestor (parent node))

    tree |> Seq.filter hasAncestor

但可能你真正想要的是一个结构，其中每个节点都存储对其子节点的引用，当你去序列化数据时，你可以从引用中找到 ID

希望这样的事情足以为您指明正确的方向：

type Node = {
    Id : int;
    Description: string;
    Children: seq<Node>
}

let myTree =
    { Id = 0; Description = "Root"; Children = 
    [
        { Id = 1; Description = "Mechanical"; Children = [] };
        { Id = 2; Description = "Electrical"; Children =         
        [
            { Id = 3; Description = "High Voltage"; Children = 
            [
                { Id = 5; Description = "HV Maintanence"; Children = [] };
                { Id = 6; Description = "City Power"; Children = [] }
            ] };
            { Id = 4; Description = "Low Voltage"; Children = 
            [
                { Id = 7; Description = "LV Wiring"; Children = [] } ;
                { Id = 8; Description = "LV Maintanence"; Children = [] }
            ] }
        ]};
    ]}

let rec getChildren (node : Node) = 
    Seq.concat [node.Children; (Seq.collect getChildren node.Children)]
于 2013-09-01T15:54:43.763 回答