XML Comparison without DTD & XSD and able to return XPath of the Diff Node

Dear All,
I am new xml concept in java.
First xml can have n of nodes and second xml can have n of the node.
The node is new to second xml when compare to first xml.It has to go change list.
1. My job is to compare 2 xml document and return the XPath of the Nodes.
2. Both xml do not have any association like XSD,DTD and XSLT.
3. I am looking for java api to find/return xpath of the node.
Please help me.
Saravanan.P

Don't know if there's a method for this in the default XML parsing interface of Java (JAXP). I never came across a method which would return an XPath expression based on a Node and a Document. Only the other way around: return the Node(s) based on a Document and an XPath expression. Maybe some Java XML framework like XStream or an Apache XML framework have this functionality. It would be possible (not really nice though) to create such a method yourself. Based on a Document and a Node as input you could iterate through the parent elements until you reach the root Node and return the path. You would need to take into account that a Node can be part of a NodeList, so you have to retrieve the index of the Node in the list too.
Ronald

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