如何查找樹視圖中根文件夾的索引？

Question

如何查找樹視圖中每個根文件夾的索引？ 假設有一個包含4個根節點的樹形視圖。他們都在同一水平上，他們都具有子節點（不顯示）：

|-a 
|-b 
|-c 
|-d 

現在讓我們假設有一個在「C」根節點的分支選擇的節點。我怎麼能得到「C」節點的索引？ （在這種情況下，它是根節點之間的第三個節點）。

因此，給定一個選定的節點，我怎麼能得到它的根節點的索引？

Answer 1

要實現你的任務，你應該找到點擊節點的父節點，然後是父節點的父節點等......所以，我們需要在這裏遞歸。

看看示例代碼（帶註釋）：

private void Form1_Load(object sender, EventArgs e) 
{ 
    //add test data on form load (you can do it on form design, too. 
    //there are 4 root nodes and each of them has one subnode. 
    //Additionally, c's first node, called 'c-1', has it's own child. 
    treeView1.Nodes.Add(new TreeNode("a")); 
    treeView1.Nodes.Add(new TreeNode("b")); 
    treeView1.Nodes.Add(new TreeNode("c")); 
    treeView1.Nodes.Add(new TreeNode("d")); 
    treeView1.Nodes[0].Nodes.Add(new TreeNode("a-1")); 
    treeView1.Nodes[1].Nodes.Add(new TreeNode("b-1")); 
    treeView1.Nodes[2].Nodes.Add(new TreeNode("c-1")); 
    treeView1.Nodes[3].Nodes.Add(new TreeNode("d-1")); 
    treeView1.Nodes[2].Nodes[0].Nodes.Add(new TreeNode("c-1-1")); 

    //expand those nodes to see things clearly 
    treeView1.ExpandAll(); 

    //subscribe to after select event. When user selects one node, treeView1_AfterSelect method will be called. 
    //this can be done on form designer too, on properties panel 
    treeView1.AfterSelect += treeView1_AfterSelect; 

} 

private void treeView1_AfterSelect(object sender, TreeViewEventArgs e) 
{ 
    //this method will be called when you select node 
    //find topmost parent by calling method FindTopMostParent and passing it selected node 
    var topmostParent = FindTopMostParent(e.Node); 

    //here's your index of topmost node (parent or grandparent or grand-grand of selcted node(
    var index = treeView1.Nodes.IndexOf(topmostParent); 

} 

private TreeNode FindTopMostParent(TreeNode node) 
{ 
    //first, we check if passed node has parent. If not, return that node, because it's the topmost one 
    //then, repeat that search for parent again and again until you find that node which does not have parent 
    while (node.Parent != null) 
     return FindTopMostParent(node.Parent); 

    //return parentless node :) 
    return node; 
}