how to create barcode in vb net 2012 Depth-first node traversal in Java

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Depth-first node traversal
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The Document Object Model: Processing Structured Documents
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Figure 4-2
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Breadth-first node traversal
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The code for processing nodes breadth first is slightly different from the depthfirst algorithm because all of the child nodes at the current level need to be processed before moving to the next level down The following code listing shows one possible example of breadth-first processing that processes each branch of a tree separately:
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processNode(Node n) { doStartNodeProcessing(n); if (nhasChildNodes()) { for (Node c = nfirstChild(); c != null; c = cnextSibling()) { // do child node processing in here } for (Node c = nfirstChild(); c != null; c = cnextSibling()) { // now move to the next level for this parent processNode(c); } } doEndNodeProcessing(n); }
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The dotted arrows show the order in which the nodes are visited In this example, the nodes are visited in the order A, B, C, D, E, F, G There is no real logic-related advantage to using breadth-first over depth-first, or vice versa, even though the breadth-first algorithm suffers from a performance
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4: Algorithms
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disadvantage because it iterates over the child nodes of a given node twice Your application should use the algorithm that is most appropriate to solving your specific traversal problem
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Position-based algorithms work by examining the locations of nodes within a given DOM document; the contents of the node are secondary (if considered at all) Two common position-based algorithms are determining whether a node has an ancestor node of a particular type and determining whether a node has a sibling of a particular type
Determining Whether a Node Has an Ancestor of a Given Type
Suppose your program needs to determine whether a node is contained within another node For example, in HTML, image tags can be contained within anchor tags, which allows them to serve as hyperlinks to other URLs, as in the following:
<a href="http://yoururlherecom"><img src="myimagejpg"></a>
If your application needs to treat a node differently depending on whether it is contained by another type of node, you can use the algorithm in the following code listing to determine if a node has a particular ancestor:
Boolean isContainedBy(Node n, DOMString nodeName) { Node testNode; testNode = ngetParentNode(); while (testNode) { if (testNodegetNodeName()equals(nodeName)) return true; testNode = testNodegetParentNode(); } return false; }
This function starts with a given node and traverses up the DOM tree, examining each node to see if the given node is contained by the node with the given name If the current parent node does not match the criteria, set the testNode to the current parent node and repeat the loop To see how this works, consider the sample document tree represented by Figure 4-3
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Figure 4-3
Sample document tree
Suppose you had a reference to the C node labeled node1 and wanted to see if the node was contained within a B node You would call the function isContainedBy() like this:
Boolean bHasBParent = isContainedBy(node1, "B");
In this case, the function would return true because node1 is contained by a B node If the same call were made on the C node labeled node2, the function would return false This example could easily be modified to see if a node was contained by a parent of a particular node type (such as Element), with particular attributes, and so on For instance, to change this example to see if a node is contained within a particular type, you would change the nodeName argument from a DOMString to a short integer representing the node type The part of the function where the test is done would be changed to check the node type instead of the node name, like this:
if (testNodegetNodeName()equals(nodeName)) return true;
Then you would call this function like this:
var bHasBParent = isContainedBy(node1, NodeELEMENT_NODE);
Determining Whether a Node Has a Sibling of a Given Type
Determining whether a node has a sibling of a particular type is similar to the previous example of determining whether it has a particular parent You simply
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