c#获取TreeView父节点列表
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c# Obtain a list of TreeView Parent nodes
提问by TK.
If I have a TreeView (myTreeview),how can I obtain a list of all the nodes that are parent nodes? i.e. nodes that have children
如果我有一个 TreeView (myTreeview)，我如何获得所有作为父节点的节点的列表？即有孩子的节点
采纳答案by ng5000
myTreeview.Nodes will give you a list of root nodes as defined by MS which basically means nodes on the root branch of the tree.
myTreeview.Nodes 会给你一个由 MS 定义的根节点列表，这基本上意味着树的根分支上的节点。
This code will build a list of root nodes with children:
此代码将构建一个包含子节点的根节点列表：
IList<TreeNode> nodesWithChildren = new List<TreeNode>();
foreach( TreeNode node in myTreeview.Nodes )
if( node.Nodes.Count > 0 ) nodesWithChildren.Add( node );
Update: and if you wanted all nodes in the TreeView that had a child regardless of how deep into the tree then use a bit of recursion, e.g.
更新：如果你想要 TreeView 中有一个孩子的所有节点，不管树有多深，那么使用一点递归，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 );
}
}
Update 2: Recursion method with only 1 foreach loop:
更新 2：只有 1 个 foreach 循环的递归方法：
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);
}
}
}
回答by user17541
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);
