Xml – What’s the best way to represent a tree with XML

data structurestreexml

I'm implementing some classes to handle the common data structures (Tree, BinaryTree, Search Binary Tree, BTree, AVL, etc). I'd like a way to store tree information in a data source and parse it to one of those classes. The simplest way is to use XML (portability is very important). The thing is that the XML documents that i make are not easy to read. They're not enough "intuitive". For example, i was doing something like this:

<?xml version="1.0" encoding="UTF-8"?>
<tree>
    <root>100</root>
    <node>
        <tree>
            <root>50</root>     
        </tree>
    </node>
    <node>
        <tree>
            <root>200</root>        
        </tree>
    </node>
</tree>

There, the tree would be something like this:

                    100
              50                200

What do you say? Do you have any other recommendation?

Best Answer

I don't see in your description any important structural difference between the root, leaves of the tree, and other nodes. So they could all have the same content model, as in this example:

<tree>
    <value>100</value>
    <tree>
        <value>50</value>
    </tree>
    <tree>
        <value>200</value>
    </tree>
</tree>

This structure is a bit less verbose, and the XML document tree is matching the tree you are trying to represent. If your values can be expressed as strings, you could even write it like that:

<tree>
    100
    <tree>50</tree>
    <tree>200</tree>
</tree>

Related Solutions

Java – What’s the best way to validate an XML file against an XSD file

The Java runtime library supports validation. Last time I checked this was the Apache Xerces parser under the covers. You should probably use a javax.xml.validation.Validator.

import javax.xml.XMLConstants;
import javax.xml.transform.Source;
import javax.xml.transform.stream.StreamSource;
import javax.xml.validation.*;
import java.net.URL;
import org.xml.sax.SAXException;
//import java.io.File; // if you use File
import java.io.IOException;
...
URL schemaFile = new URL("http://host:port/filename.xsd");
// webapp example xsd: 
// URL schemaFile = new URL("http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd");
// local file example:
// File schemaFile = new File("/location/to/localfile.xsd"); // etc.
Source xmlFile = new StreamSource(new File("web.xml"));
SchemaFactory schemaFactory = SchemaFactory
    .newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
try {
  Schema schema = schemaFactory.newSchema(schemaFile);
  Validator validator = schema.newValidator();
  validator.validate(xmlFile);
  System.out.println(xmlFile.getSystemId() + " is valid");
} catch (SAXException e) {
  System.out.println(xmlFile.getSystemId() + " is NOT valid reason:" + e);
} catch (IOException e) {}

The schema factory constant is the string http://www.w3.org/2001/XMLSchema which defines XSDs. The above code validates a WAR deployment descriptor against the URL http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd but you could just as easily validate against a local file.

You should not use the DOMParser to validate a document (unless your goal is to create a document object model anyway). This will start creating DOM objects as it parses the document - wasteful if you aren't going to use them.

Xml – What’s the difference between an element and a node in XML

The Node object is the primary data type for the entire DOM.

A node can be an element node, an attribute node, a text node, or any other of the node types explained in the "Node types" chapter.

An XML element is everything from (including) the element's start tag to (including) the element's end tag.

Best Answer

Related Solutions

Java – What’s the best way to validate an XML file against an XSD file

Xml – What’s the difference between an element and a node in XML

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