Count tree root nodes

[WojTec]

Today I need to know how many root nodes (I mean without parent node) has my tree. I wrote this:

Delphi-Quellcode:

function CountTreeNodes(ATree: TTreeView; ALevel: Word): Integer;
var
  I: integer;
begin
  Result := 0;

  for I := 0 to ATree.Items.Count - 1 do
  begin
    if ALevel = ATree.Items[I].Level then
      Inc(Result)
    ;
  end;
end;

Count := CountTreeNodes(Tree, 0);

This is working, but is quite lame IMO - function will be slow and sloow and slooow if many nodes on tree. Is possible to do it in pro mode?

[Uwe Raabe]

This should be slightly faster:

Delphi-Quellcode:

function CountRootNodes(ATree: TTreeView): Integer;
var
  node: TTreeNode;
begin
  Result := 0;

  node := ATree.Items.GetFirstNode;
  while node <> nil do begin
    Inc(Result);
    node := node.GetNextSibling;
  end;
end;

[Sir Rufo]

You are right that this is very slow and the documentation also stated this TTreeView.Items

http://docwiki.embarcadero.com/Libraries/XE6/en/Vcl.ComCtrls.TTreeView.Items

Note: Accessing tree view items by index can be time-intensive, particularly when the tree view contains many items. For optimal performance, try to design your application so that it has as few dependencies on the tree view's item index as possible.

But there are some methods to walk through the tree:

Get the very first node by calling MyNode := MyTreeView.Items[0]; .
Walk through the tree until you get the desired level with TTreeNode.Next

Delphi-Quellcode:

while Assigned( MyNode ) and ( MyNode.Level <> ALevel ) do
  MyNode := MyNode.GetNext;

Once you got the node at the desired level you can step through all nodes on this level with TTreeNode.GetNextSibling

Delphi-Quellcode:

while Assigned( MyNode ) do
  MyNode := MyNode.GetNextSibling;

[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