c＃获取TreeView父节点的列表

Question

如果我有一个TreeView（myTreeview），如何获取所有父节点的节点列表？即有子女的节点

Answer 1

myTreeview.Nodes将为您提供MS定义的根节点列表，这些节点基本上是指树的根分支上的节点。

该代码将建立根节点的儿童名单：

IList<TreeNode> nodesWithChildren = new List<TreeNode>(); 
    foreach(TreeNode node in myTreeview.Nodes) 
     if(node.Nodes.Count > 0) nodesWithChildren.Add(node); 

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更新：如果你想在那有一个孩子TreeView中的所有节点，无论有多深成树，然后使用一点递归，例如

private static IList<TreeNode> BuildParentNodeList(TreeView treeView) 
{ 
    IList<TreeNode> nodesWithChildren = new List<TreeNode>(); 

    foreach(TreeNode node in treeView.Nodes ) 
     AddParentNodes(nodesWithChildren, node); 

    return nodesWithChildren; 
} 

private static void AddParentNodes(IList<TreeNode> nodesWithChildren, TreeNode parentNode) 
{ 
    if (parentNode.Nodes.Count > 0) 
    { 
     nodesWithChildren.Add(parentNode); 
     foreach(TreeNode node in parentNode.Nodes) 
      AddParentNodes(nodesWithChildren, node); 
    } 
} 

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更新2：递归方法只有1 foreach循环：

private static IList<TreeNode> BuildParentNodeList(TreeView treeView) 
{ 
    IList<TreeNode> nodesWithChildren = new List<TreeNode>(); 
    AddParentNodes(nodesWithChildren, treeView.Nodes); 
    return nodesWithChildren; 
} 

private static void AddParentNodes(IList<TreeNode> nodesWithChildren, TreeNodeCollection parentNodes) 
{ 
    foreach (TreeNode node in parentNodes) 
    { 
     if (node.Nodes.Count > 0) 
     { 
      nodesWithChildren.Add(node); 
      AddParentNodes(nodesWithChildren, node.Nodes); 
     } 
    } 
} 

Answer 2

private void AddNonLeafNodes(List<TreeNode> nonLeafNodes, TreeNodeCollection nodes) 
{ 
    foreach(TreeNode node in nodes) 
    { 
     if(node.Nodes.Count > 0) 
     { 
      nonLeafNodes.Add(node); 
      AddNonLeafNodes(nonLeafNodes,node.Nodes); 
     } 
    } 
} 

List<TreeNode> nonLeafNodes = new List<TreeNode>(); 
AddNonLeafNodes(nonLeafNodes,treeView1.Nodes);