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！