Get a list of all tree nodes (in all levels) in TreeView Controls

Question

How can I get a list of all tree nodes (in all levels) in a TreeView control?

Answer 1

You can use two recursive extension methods. You can either call myTreeView.GetAllNodes() or myTreeNode.GetAllNodes():

public static List<TreeNode> GetAllNodes(this TreeView _self)
{
    List<TreeNode> result = new List<TreeNode>();
    foreach (TreeNode child in _self.Nodes)
    {
        result.AddRange(child.GetAllNodes());
    }
    return result;
}

public static List<TreeNode> GetAllNodes(this TreeNode _self)
{
    List<TreeNode> result = new List<TreeNode>();
    result.Add(_self);
    foreach (TreeNode child in _self.Nodes)
    {
        result.AddRange(child.GetAllNodes());
    }
    return result;
}

Answer 2

If you don't need the node Key to be unique, just set all of the node keys to an empty string (""), then you can do a Treeview1.Nodes.Find("", true); to return all nodes within a TreeView.

Answer 3

Update to Krumelur's answer (replace 2 first lines of his/her solution with this):

foreach ( var node in oYourTreeView.Nodes )
{
    PrintNodesRecursive( node );
}

Answer 4

Because TreeView has many levels, do recursive function:

   public void AddNodeAndChildNodesToList(TreeNode node)
    {
        listBox1.Items.Add(node.Text);    // Adding current nodename to ListBox     

        foreach (TreeNode actualNode in node.Nodes)
        {
            AddNodeAndChildNodesToList(actualNode); // recursive call
        }
    }

Than call this function for all first level Nodes in TreeView:

    foreach (TreeNode actualNode in treeView1.Nodes)         // Begin with Nodes from TreeView
    {
         AddNodeAndChildNodesToList(actualNode);
    }

Code is from site C# TreeView

Answer 5

I think my solution is more elegant, it uses generics (because TreeView can store each kind of objects derived from TreeNode) and has a single function recursively called. It should be straightforward also convert this as extension.

    List<T> EnumerateAllTreeNodes<T>(TreeView tree, T parentNode = null) where T : TreeNode
    {
        if (parentNode != null && parentNode.Nodes.Count == 0)
            return new List<T>() { };

        TreeNodeCollection nodes = parentNode != null ? parentNode.Nodes : tree.Nodes;
        List<T> childList = nodes.Cast<T>().ToList();

        List<T> result = new List<T>(1024); //Preallocate space for children
        result.AddRange(childList); //Level first

        //Recursion on each child node
        childList.ForEach(n => result.AddRange(EnumerateAllTreeNodes(tree,n)));

        return result;
    }

The usage is straightforward, just call:

    List<MyNodeType> allnodes = EnumerateAllTreeNodes<MyNodeType>(tree);

Answer 6

Lazy LINQ approach, just in case you're looking for something like this:

private void EnumerateAllNodes()
{
    TreeView myTree = ...;

    var allNodes = myTree.Nodes
        .Cast<TreeNode>()
        .SelectMany(GetNodeBranch);

    foreach (var treeNode in allNodes)
    {
        // Do something
    }
}

private IEnumerable<TreeNode> GetNodeBranch(TreeNode node)
{
    yield return node;

    foreach (TreeNode child in node.Nodes)
        foreach (var childChild in GetNodeBranch(child))
            yield return childChild;
}