DevExpress实现TreeList向上递归获取公共父节点的方法

/// <summary>
/// 向上递归，获取符合条件的父节点
/// </summary>
/// <param name="node">需要向上递归的节点</param>
/// <param name="conditionHanlder">判断条件【委托】</param>
/// <returns>符合条件的节点【TreeListNode】</returns>
public static TreeListNode GetParentNode(this TreeListNode node, Predicate<TreeListNode> conditionHanlder)
{
  TreeListNode _parentNode = node.ParentNode;//获取上一级父节点
  TreeListNode _conditonNode = null;
  if (_parentNode != null)
  {
 if (conditionHanlder(_parentNode))//判断上一级父节点是否符合要求
 {
   _conditonNode = _parentNode;
 }
 if (_conditonNode == null)//若没有找到符合要求的节点，递归继续
   _conditonNode = GetParentNode(_parentNode, conditionHanlder);
  }
  return _conditonNode;
}
/// <summary>
/// 向上递归节点
/// </summary>
/// <param name="node">需要向上递归的节点</param>
/// <param name="conditionHanlder">委托，返回fasle跳出递归；返回true继续递归；</param>
public static void UpwardRecursiveNode(this TreeListNode node, Predicate<TreeListNode> conditionHanlder)
{
  TreeListNode _parentNode = node.ParentNode;
  if (_parentNode != null)
  {
 if (conditionHanlder(_parentNode))
 {
   UpwardRecursiveNode(_parentNode, conditionHanlder);
 }
  }
}
/// <summary>
/// 向上递归，获取符合条件的节点的公共父节点
/// </summary>
/// <param name="node">操作节点</param>
/// <param name="checkHanlder">委托</param>
/// <returns>符合条件的节点</returns>
public static TreeListNode 
GetPublicParentNode

(this TreeListNode node, Predicate<TreeListNode> checkHanlder)
{
  TreeListNode _publicPNode = null;
  TreeListNode _findNode = node.GetParentNode(checkHanlder);//先获取到条件判断的自身父节点
  if (_findNode != null)
  {
 //开始向上递归
 UpwardRecursiveNode(_findNode, n =>
 {
   TreeListNode _curpublicNode = n.ParentNode;//获取当前向上递归的父节点
   if (_curpublicNode != null)
   {
 if (_curpublicNode.Nodes.Count > 1)//若有多个子节点，则是公共父节点
 {
   _publicPNode = _curpublicNode;
   return false;//跳出递归
 }
   }
   return true;//继续递归
 });
  }
  return _publicPNode;
}