Wie kann Knoten eines Baums unabhängig von ihrer Position abfragen?
public
class
TreeNode
{
string
Id {
get
;
set
; }
int
Data {
List<TreeNode> Children {
private
TreeNode(
id,
data)
:
this
(id, data,
new
TreeNode[0])
{ }
node,
data,
params
TreeNode[] children)
Id = node;
Data = data;
Children =
List<TreeNode>(children);
}
TreeNode rootNode =
"node-0"
, 35,
"node-1"
, 17,
"node-2"
, 20)),
"node-3"
, 19),
"node-4"
, 5,
"node-5"
, 25),
"node-6"
, 40)));
↑ Zurück zum Anfang
static
LinqTreeExtension
IEnumerable<T> SelectNestedChildren<T>
(
IEnumerable<T> source, Func<T, IEnumerable<T>> selector)
foreach
(T item
in
source)
yield
return
item;
(T subItem
SelectNestedChildren(selector(item), selector))
subItem;
var flattenedTreeWithoutRoot
= rootNode.Children
.SelectNestedChildren(t => t.Children).ToList();
var flattenedTreeIncludingRoot
=
TreeNode[] {rootNode }
var result
TreeNode[] { rootNode }
.SelectNestedChildren(t => t.Children)
.Where(t => t.Data > 30).ToList();
var databaseResult = context.TreeNodes
var xmlElements = XDocument.Load(
"myfile.xml"
).Descendants();
var result = xmlElements.Where(x => ...);
Carsten Siemens edited Revision 4. Comment: Fixed typo
Carsten Siemens edited Revision 3. Comment: Fixed misspelling
Carsten Siemens edited Revision 2. Comment: Fixed link "Zurück zum Anfang"
Carsten Siemens edited Revision 1. Comment: Fixed links "Zurück zum Anfang"
Carsten Siemens edited Original. Comment: Fixed language tag. Added "(de-DE)" to title.