Count tree root nodes

[jaenicke]

If performance is an issue for you, why don't you use the virtual trees? They are faster by orders of magnitude...
http://www.jam-software.com/virtual-treeview/

And for your problem you have VirtualStringTree1.ChildCount[CurrentNode] .
This property only fetches the appropriate value, because this value is already known, so there is no need for counting or anything. Faster is impossible.

[Sir Rufo]

Zitat von jaenicke:

And for your problem you have VirtualStringTree1.ChildCount[CurrentNode] .
This property only fetches the appropriate value, because this value is already known, so there is no need for counting or anything. Faster is impossible.

It is fast, because it did not count all wanted nodes

Code:

+-n1
  +-n2
  +-n3
+-n4
  +-n5
  +-n6

Counting all nodes at level 1 should return 4 and not 2

[WojTec]

I used @Uwe Raabe and @Sir Rufo method, also thanks for explanation