问题
I am working on a multi-level marketing (binary) which looks like this:
(but the binary tree is not required to be perfect. A node can have 0-2 child)
My problem is the data that I fetch from the database is flat list.
Notice that I am using hierarchyid (sql server 2014)
Basically the TextNode column is like a breadcrumb.
every slash / represents a level.
If I have TextNode of /1/ as root. then every node that starts with /1/ belongs to that root which are /1/, /1/1/ and /1/1/1/ (the root node is included which will be the level 0)
I've tried the accepted answer in this question but its not working.
How can I transform the flatlist to a Binary Tree so that I can easily traverse and display it on a screen?
Im using C#, ASP MVC 5, SQL Server 2014 if it matters.
回答1:
I implement exactly this code According to Alex implementation but as is mentioned in some case it didn't work correctly .. have a look to my Image and my code (which copied from Alex post) [data in the database are correct but in tree view seems some problems ]
public class Row : IRow<string>
{
public string TextNode { get; }
public string Value { get; }
public long Id { get; }
public string FIN { get; }
public Row(string textNode, string userName, long id, string fin)
{
FIN = fin;
Id = id;
TextNode = textNode;
Value = userName;
}
}
public interface IRow<out T>
{
string TextNode { get; }
long Id { get; }
string FIN { get; }
T Value { get; }
}
public class TreeNode<T>
{
private struct NodeDescriptor
{
public int Level { get; }
public int ParentIndex { get; }
public NodeDescriptor(IRow<T> row)
{
var split = row.TextNode.Split(new[] { "/" }, StringSplitOptions.RemoveEmptyEntries);
Level = split.Length;
ParentIndex = split.Length > 1 ? int.Parse(split[split.Length - 2]) - 1 : 0;
}
}
public T title { get; }
public long Id { get; }
public string FIN { get; }
public List<TreeNode<T>> children { get; }
private TreeNode(T value, long id, string fin)
{
Id = id;
FIN = fin;
title = value;
children = new List<TreeNode<T>>();
}
public static TreeNode<T> Parse(IReadOnlyList<IRow<T>> rows)
{
if (rows.Count == 0)
return null;
var result = new TreeNode<T>(rows[0].Value, rows[0].Id, rows[0].FIN);
FillParents(new[] { result }, rows, 1, 1);
return result;
}
private static void FillParents(IList<TreeNode<T>> parents, IReadOnlyList<IRow<T>> rows, int index, int currentLevel)
{
var result = new List<TreeNode<T>>();
for (int i = index; i < rows.Count; i++)
{
var descriptor = new NodeDescriptor(rows[i]);
if (descriptor.Level != currentLevel)
{
FillParents(result, rows, i, descriptor.Level);
return;
}
var treeNode = new TreeNode<T>(rows[i].Value, rows[i].Id, rows[i].FIN);
parents[descriptor.ParentIndex].children.Add(treeNode);
result.Add(treeNode);
}
}
}
g
this is also my JSON output for more information :
{"title":"Earth","Id":32,"FIN":"FIN","children":[{"title":"Europe","Id":33,"FIN":"FIN001","children":[{"title":"France","Id":35,"FIN":"FIN001001","children":[{"title":"Paris","Id":36,"FIN":"FIN001001001","children":[]},{"title":"Brasilia","Id":41,"FIN":"FIN002001001","children":[]},{"title":"Bahia","Id":42,"FIN":"FIN002001002","children":[]}]},{"title":"Spain","Id":38,"FIN":"FIN001002","children":[{"title":"Madrid","Id":37,"FIN":"FIN001002001","children":[{"title":"Salvador","Id":43,"FIN":"FIN002001002001","children":[]}]}]},{"title":"Italy","Id":45,"FIN":"FIN001003","children":[]},{"title":"Germany","Id":48,"FIN":"FIN001004","children":[]},{"title":"test","Id":10049,"FIN":"FIN001005","children":[]}]},{"title":"South America","Id":34,"FIN":"FIN002","children":[{"title":"Brazil","Id":40,"FIN":"FIN002001","children":[{"title":"Morano","Id":47,"FIN":"FIN001003001","children":[]}]}]},{"title":"Antarctica","Id":39,"FIN":"FIN003","children":[{"title":"McMurdo Station","Id":44,"FIN":"FIN003001","children":[]}]}]}
回答2:
Here is a very simple implementation (assuming that Nodes are in the right order), which may be enhanced in multiple ways
public interface IRow<out T>
{
string TextNode { get; }
T Value { get; }
}
public class TreeNode<T>
{
private struct NodeDescriptor
{
public int Level { get; }
public int ParentIndex { get; }
public NodeDescriptor(IRow<T> row)
{
var split = row.TextNode.Split(new [] {"/"}, StringSplitOptions.RemoveEmptyEntries);
Level = split.Length;
ParentIndex = split.Length > 1 ? int.Parse(split[split.Length - 2]) - 1 : 0;
}
}
public T Value { get; }
public List<TreeNode<T>> Descendants { get; }
private TreeNode(T value)
{
Value = value;
Descendants = new List<TreeNode<T>>();
}
public static TreeNode<T> Parse(IReadOnlyList<IRow<T>> rows)
{
if (rows.Count == 0)
return null;
var result = new TreeNode<T>(rows[0].Value);
FillParents(new[] {result}, rows, 1, 1);
return result;
}
private static void FillParents(IList<TreeNode<T>> parents, IReadOnlyList<IRow<T>> rows, int index, int currentLevel)
{
var result = new List<TreeNode<T>>();
for (int i = index; i < rows.Count; i++)
{
var descriptor = new NodeDescriptor(rows[i]);
if (descriptor.Level != currentLevel)
{
FillParents(result, rows, i, descriptor.Level);
return;
}
var treeNode = new TreeNode<T>(rows[i].Value);
parents[descriptor.ParentIndex].Descendants.Add(treeNode);
result.Add(treeNode);
}
}
}
Sample usage:
public class Row : IRow<string>
{
public string TextNode { get; }
public string Value { get; }
public Row(string textNode, string userName)
{
TextNode = textNode;
Value = userName;
}
}
class Program
{
static void Main(string[] args)
{
IRow<string>[] rows =
{
new Row("/", "Ahmed"),
new Row("/1/", "Saeed"),
new Row("/2/", "Amjid"),
new Row("/1/1/", "Noura"),
new Row("/2/1/", "Noura01"),
new Row("/2/2/", "Reem01"),
new Row("/1/1/1", "Under_noura")
};
var tree = TreeNode<string>.Parse(rows);
PrintTree(tree);
}
private static void PrintTree<T>(TreeNode<T> tree, int level = 0)
{
string prefix = new string('-', level*2);
Console.WriteLine("{0}{1}", prefix, tree.Value);
foreach (var node in tree.Descendants)
{
PrintTree(node, level + 1);
}
}
}
来源：https://stackoverflow.com/questions/32044123/how-to-transform-list-of-hierarchyid-into-a-binary-tree