[xsl] Node Position & Relationship!

["Adam J Knight" <adam@xxxxxxxxxxxxxxxxx>]

Given the following xml structure, I want to create an xsl if element that
tests the current node to see if it is a level 0 tree_node element.

<?xml version="1.0"?>
<tree>
  <tree_node id="7" value="Test">
      <tree_node id="8" value="Test Sub"/>
      <tree_node id="9" value="Test Sub One">
          <tree_node id="10" value="Test Sub Two"/>
    </tree_node>
  </tree_node>
</tree>


Here is my attached, pretty sad!!!!!

<xsl:if test="{count(self::*)=1}">
  	  <xsl:apply-templates select="tree_node"/>
</xsl:if

Can someone give me the correct way to achieve this.
Also how can I found out if a particular node is a child, ancestor or peer
Of any given node.

Thanks to anyone who response, muchly appreciated.

Cheers,
Adam
 
NB: "Pray as if everything depended upon God and work as if everything
depended upon man."