SKSenthil Kumaran Rajan Syncfusion Team June 10, 2014 10:30 AM UTC
Thankyou for using
We would like to let you know that your requirement in
"Treeviewadv" can be achieved by creating another list. In the sample
provided, we have found that only a single
list has been maintained due to the nodes which have been checked into that list. We have prepared a sample to achieve
your query. Please have the sample attached below.
Code Snippet[C#] : To get a child node in the list.
List<String> child = new List<String>();
for (int n = 0; n < treeViewAdv1.Nodes.Count; n++)
(TreeNodeAdv childnode in treeViewAdv1.Nodes[n].Nodes)