Question

我有一个具有以下结构的课程

public class Menu
{
    public Menu()
    {
        ChildMenus = new List<Menu>();
    }

    public string MenuName { get; set; }
    public int Id { get; set; }
    public int Level { get; set; }
    public int Order { get; set; }
    public string RefCode { get; set; }
    public List<Menu> ChildMenus { get; set; }
    public int ParentId { get; set; }
}

菜单对象的集合，如：

public class MenuManager
{
    public List<Menu> Menucollection;

    public MenuManager()
    {
        Menucollection = new List<Menu>
        {
            new Menu {MenuName = "Products", Id = 1, ParentId = 0, Level = 1, RefCode = "0", Order = 1},
            new Menu {MenuName = "Sales", Id = 2, ParentId = 0, Level = 1, RefCode = "0", Order = 2},
            new Menu {MenuName = "Contacts", Id = 3, ParentId = 0, Level = 1, RefCode = "0", Order = 3},
            new Menu {MenuName = "Mobiles", Id = 4, ParentId = 1, Level = 2, RefCode = "0", Order = 1},
            new Menu {MenuName = "Computers", Id = 5, ParentId = 1, Level = 2, RefCode = "0", Order = 2},
            new Menu {MenuName = "Local Sales", Id = 6, ParentId = 2, Level = 2, RefCode = "0", Order = 1},
            new Menu {MenuName = "Forgin Sales", Id = 7, ParentId = 2, Level = 2, RefCode = "0", Order = 2},
            new Menu {MenuName = "Email", Id = 8, ParentId = 3, Level = 2, RefCode = "100", Order = 1},
            new Menu {MenuName = "Nokia", Id = 9, ParentId = 4, Level = 3, RefCode = "110", Order = 1},
            new Menu {MenuName = "LG", Id = 10, ParentId = 4, Level = 3, RefCode = "111", Order = 2},
            new Menu {MenuName = "Accer", Id = 11, ParentId = 5, Level = 3, RefCode = "210", Order = 1},
            new Menu {MenuName = "Kerala", Id = 9, ParentId = 6, Level = 3, RefCode = "110", Order = 1}
        };
    }

    public void BuildMenu()
    {
    }
}

我需要从集合中构建一个菜单结构，条件如下：

如果出现以下情况，必须删除菜单项：

如果没有任何孩子。
如果其重新编码等于0。

因此，基于输入的预期结果应为：

Product    Sales         Contact
Mobile     Local Sales   Email
Nokia      Kerala
LG
Computer 
Accer

Answer 1

首先，与之前的评论相反，不需要递归来处理树。它通常是最简单，最简单的方法（顺便提一下，我提供的方法） - 但不是必需的（并且有时可以避免递归的性能原因）。

注意，扩展方法是基于我从这里发送的剪切我：Searching a tree using LINQ

            var filteredMenus = from m in Menucollection
                                where m.RefCode != "0" || Menucollection.Any(m1=>m1.ParentId == m.Id)
                                select m;

            foreach(var m in filteredMenus)
            {
                m.ChildMenus = filteredMenus.Where(m1=>m1.ParentId == m.Id).ToList();
            }

            var expanded = from m in filteredMenus
                       where m.ParentId == 0
                       select new { Menu = m, Descendants = m.AsBreadthFirstEnumerable(m1 => m1.ChildMenus) };

            foreach (var m in expanded)
            {
                System.Diagnostics.Debug.WriteLine("Menu: " + m.Menu.MenuName);
                foreach (var m1 in m.Descendants)
                {
                    System.Diagnostics.Debug.WriteLine("- " + m1.MenuName);
                }
            }

然后在一些静态类中：

    public static IEnumerable<T> AsBreadthFirstEnumerable<T>(this T head, Func<T, IEnumerable<T>> childrenFunc)
    {
        yield return head;

        var last = head;
        foreach (var node in AsBreadthFirstEnumerable(head, childrenFunc))
        {
            foreach (var child in childrenFunc(node))
            {
                yield return child;
                last = child;
            }
            if (last.Equals(node)) yield break;
        }
    }

它不是纯粹的LINQ - 但它应该给你一些起点。

如何使用linq（非二叉树）构建树节点关系

1 个答案: