Current test site is here:  http://cato.maticeinteractive.com

If you are on top level page that has a sub-menu, I want the sub-menu to show.
If you are on a sub-menu page that has a sub-sub-menu, I want the sub-sub-menu to show.

Obviously I can't seem to get my xslt correct for this to happen.  Here's what I've got (my base code that works). I've tried all sorts of things but none seem to do what I want.

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
     <xsl:output method="html"/>
    <xsl:param name="menuType"></xsl:param>
     <xsl:template match="/*">
           <xsl:apply-templates select="root" />
     </xsl:template>
     <xsl:template match="root">
        <ul>
            <xsl:attribute name="class">menu<xsl:text>&#32;</xsl:text><xsl:value-of select="$menuType"/></xsl:attribute>        
               <xsl:apply-templates select="node" />
          </ul>
     </xsl:template>
     <xsl:template match="node">
          <li>
               <xsl:attribute name="class">
                    <xsl:if test="@breadcrumb = 1"><xsl:text>current
                </xsl:text></xsl:if>
               </xsl:attribute>     
               <xsl:choose>
                    <xsl:when test="@selected = 1">
                         <xsl:apply-templates select="node" />
                    </xsl:when>
                    <xsl:when test="@enabled = 1">
                         <a href="{@url}">
                              <xsl:value-of select="@text" />
                         </a>
                    </xsl:when>
                    <xsl:otherwise>
                         <xsl:value-of select="@text" />
                    </xsl:otherwise>
               </xsl:choose>
               <xsl:if test="node">
                    <ul>
                         <xsl:apply-templates select="node" />
                    </ul>
               </xsl:if>
          </li>
     </xsl:template>
</xsl:stylesheet>

Thanks for any help!