递归获取节点是很多程序项目中常见的技巧。本文就以实例展示了DevExpress获取节点下可视区域子节点集合的实现方法。分享给大家供参考之用，具体方法如下：
关键部分代码如下：
/// <summary>
/// 向下递归TreeListNode节点
/// </summary>
/// <param name="node">需要向下递归的节点</param>
/// <param name="conditionHanlder">委托</param>
public static void DownRecursiveNode(this TreeListNode node, Action<TreeListNode> conditionHanlder)
{
foreach (TreeListNode _childNode in node.Nodes)
{
conditionHanlder(_childNode);
DownRecursiveNode(_childNode, conditionHanlder);
}
}
/// <summary>
/// 获取节点下可视区域子节点集合
/// </summary>
/// <param name="node">需要获取可见子节点的节点</param>
/// <param name="conditonHanlder">条件委托</param>
/// <returns>可见子节点集合</returns>
public static List<TreeListNode> GetVisibleChildNodes(this TreeListNode node, Predicate<TreeListNode> conditonHanlder)
{
List<TreeListNode> _visibleChildNodes = new List<TreeListNode>();
TreeList _tree = node.TreeList;
DownRecursiveNode(node, n =>
{
RowInfo _rowInfo = _tree.ViewInfo.RowsInfo[n];
if (_rowInfo != null)
{
if (conditonHanlder(n))
{
_visibleChildNodes.Add(n);
}
}
});
return _visibleChildNodes;
}
/// <summary>
/// 获取节点下可视区域子节点集合
/// </summary>
/// <param name="node">需要获取可见子节点的节点</param>
/// <returns>可见子节点集合</returns>
public static List<TreeListNode> GetVisibleChildNodes(this TreeListNode node)
{
return GetVisibleChildNodes(node, n => 1 == 1);
}
希望本文所述方法对大家的C#程序设计能有所帮助！
