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JTree with nodes generated at run-time

My program builds a tree based on input data. Before running the program, there is no way of knowing how many nodes should be created and where the position of the nodes (under which parent) should be.
Using JTree, we can add nodes easily if we, beforehand, are aware of the structure.
E.g.
//create the root node
DefaultMutableTreeNode root = new DefaultMutableTreeNode("Root");
/create the child nodes
DefaultMutableTreeNode vegetableNode = new DefaultMutableTreeNode("Vegetables");
vegetableNode.add(new DefaultMutableTreeNode("Capsicum"));
vegetableNode.add(new DefaultMutableTreeNode("Carrot"));
vegetableNode.add(new DefaultMutableTreeNode("Tomato"));
vegetableNode.add(new DefaultMutableTreeNode("Potato"));
root.add(vegetableNode);
But my tree, once the root node is specified, is incrementally built. So I'd like my JTree to add a node under a specific parent at any time.
Please note that the tree will also be built inside a recursive method. Which means the main JTree object should be created outside this method.
The tree may have more than one layer, which means the path from the node to a leaf may require e.g. 10 jumps.
What is the best way of adding a node to a JTree at run-time which only knows about its parent?
A caveat here is when two different parents have the same name e.g. if node apple should go under a node called fruit but fruit is found in two different paths
root -> aaa -> bbb -> ccc -> fruit
root -> aaa -> fff -> ggg -> hhh -> fruit

You should look into providing your own tree model then.
Quoting the Oracle tutorial for JTree:
One of the ways you can lazily load children of a Tree is by utilizing the TreeWillExpandListener interface. For example, you can declare and load root, grandparent and parent of a Tree along with the application as shown in the following code:
Given the comments by the OP: the DefaultMutableTreeNode does not support "names". That would mean: you need to implement that yourself, in a super simply approach, probably like:
class MyTreeNode extends DefaultMutableTreeNode {
private final String name;
MyTreeNode(String name, ... whatever ) {
...
SOME_NODE_REGISTRY.put(name, this);
}
Then you need to provide that registry, probably as some sort of singleton map instance. In other words: you have to write code that A) enables you to store nodes by name to then B) identify/find nodes by name.

threaded binary tree

Hello guys i am supposed to write the ThreadedNode() class, but im haveing a few problems with it.
I understand that a threaded binary tree of a binary tree is obtained by setting every null left child to the predecessor of the node in the inorder traversal and every null right child
to the successor of the node in the inorder traversal.
however i have my problem starts with the constructor
// thread the binary tree when you are given the root
public ThreadedNode( BinaryNode root)
i know it receives a binaryNode and i have to make it a threaded tree, but i how do create the new threaded tree?

A common way to create threaded binary trees is with a fake head. This makes the single node trees simpler to understand and the constructor more straightforward.
Thus your constructor would probably look like:
public class ThreadedNode {
private BinaryNode head;
public ThreadedNode(BinaryNode root) {
head = new BinaryNode();
root.makeThreaded();
root.setRight(head);
head.setRight(root);
}
}
Remember that later you need to account for this fake head in insert, delete, etc.

How to populate JTree dynamically?

I am working on a project that requires the use of a tree data structure. Having done some researches I found that Java JTree would be of great use to my project, however I stumbled upon a problem that I spent a week fixing but to no avail.
Here's the problem, in order to create a new node, a DefaultMutableTreeNode has to be instantiated and I'm not sure how that can be done in a loop. Normally when we want to create a new node in JTree, we would firstly declare the nodes in the following way:
DefaultMutableTreeNode parent = new DefaultMutableTreeNode("This is parent node.");
DefaultMutableTreeNode child = new DefaultMutableTreeNode("This is child node.");
Then, in order to add/link child node to parent node, we would do the following:
parent.add(child);
I have two arraylists containing the parent nodes and child nodes, they correspond to each other in a parent-child relationship, meaning that arraylistParent.get(x) will always be the parent of arraylistChild.get(x).
I was thinking that by using a for loop, I could do:
for (int x = 0; x < arraylistParent.size(); x++){
parent.add(new DefaultMutableTreeNode(arraylistChild.get(x)));
}
This could only work in a flat hierarchy tree, which obviously isn't the case for me. I will have different parent nodes in arrayListParent which I need to check before adding the child nodes, but once again not all child nodes have the same, single parent node. My arraylists probably contain something like this:
arraylistParent = [root, p1, p2, p2, p3, p1]
arraylistChild = [p1, p2, p5, p3, p4, p5]
and I want to generate the tree structure like this:
root
..p1
..p2
..p5
..p3
..p4
..p5
Obviously I can throw a few if loops inside to check whether parent.getUserObject() is the same as arraylistParent.get(x) but only those that match the string "This is parent node" will be checked and added.
If there's a different, non-existent parent found in the loop, a new parent needs to be created and this is the tricky part I don't know how to solve, as I have no idea how to have the loop automatically create a new instance of parent for the child node.

Let us assume your node-datums are Strings (that is, the contents of your ArrayLists are Strings). The explanation will work with any other Java object, though. We will call these contents 'nodes', as distinct from 'tree nodes'.
You can create a HashMap<String, DefautMutableTreeNode> m, so that m.get(node) will return the corresponding TreeNode.
Now, you have to iterate over the parents' arraylist. At each position, you will have a parent and a child. Look them up in the map. If the parent TreeNode does not exist, you will first have to create it. If the child TreeNode does not exist, the same will apply. You should make sure that the map is updated after creating any TreeNodes. Finally, you will mark the child as a child of the parent.
At the end of this algorithm, you should look through all the nodes in the child array. The one without any parents is the root. This is the root of your tree.

How to build in Java a Weighted Directed Acyclic Graph

I did a search on similar topics, but the answers are too vague for my level of understanding and comprehension, and I don't think they're specific enough to my question.
Similar threads:
Tree (directed acyclic graph) implementation
Representing a DAG (directed acyclic graph)
Question:
I have formatted a text file which contains data of the following format...
Example dataset:
GO:0000109#is_a: GO:0000110#is_a: GO:0000111#is_a: GO:0000112#is_a: GO:0000113#is_a: GO:0070312#is_a: GO:0070522#is_a: GO:0070912#is_a: GO:0070913#is_a: GO:0071942#part_of: GO:0008622
GO:0000112#part_of: GO:0000442
GO:0000118#is_a: GO:0016581#is_a: GO:0034967#is_a: GO:0070210#is_a: GO:0070211#is_a: GO:0070822#is_a: GO:0070823#is_a: GO:0070824
GO:0000120#is_a: GO:0000500#is_a: GO:0005668#is_a: GO:0070860
GO:0000123#is_a: GO:0005671#is_a: GO:0043189#is_a: GO:0070461#is_a: GO:0070775#is_a: GO:0072487
GO:0000126#is_a: GO:0034732#is_a: GO:0034733
GO:0000127#part_of: GO:0034734#part_of: GO:0034735
GO:0000133#is_a: GO:0031560#is_a: GO:0031561#is_a: GO:0031562#is_a: GO:0031563#part_of: GO:0031500
GO:0000137#part_of: GO:0000136
I'm looking to construct a weighted directed DAG from this data (the above is just a snippet). The whole dataset of 106kb is here: Source
--------------------------------------------------
Taking into consideration line-by-line, the data of each line is explained as follows...
First line as an example:
GO:0000109#is_a: GO:0000110#is_a: GO:0000111#is_a: GO:0000112#is_a: GO:0000113#is_a: GO:0070312#is_a: GO:0070522#is_a: GO:0070912#is_a: GO:0070913#is_a: GO:0071942#part_of: GO:0008622
'#' is the delimeter/tokenizer for the line data.
The First term, GO:0000109 is the node name.
The subsequent terms of is_a: GO:xxxxxxx OR part_of: GO:xxxxxxx are the nodes which are connected to GO:0000109.
Some of the subsequent terms have connections too, as depicted in the dataset.
When it is is_a, the weight of the edge is 0.8.
When it is part_of, the weight of the edge is 0.6.
--------------------------------------------------
I have Google-d on how DAGs are, and I understand the concept. However, I still have no idea how to put it into code. I'm using Java.
From my understanding, a graph generally consists of nodes and arcs. Does this graph require an adjacency list to determine the direction of the connection? If so, I'm not sure how to combine the graph and adjacency list to communicate with each other.
After constructing the graph, my secondary goal is to find out the degree of each node from the root node. There is a root node in the dataset.
For illustration, I have drawn out a sample of the connection of the first line of data below:
Image Link
I hope you guys understand what I'm trying to achieve here. Thanks for looking through my problem. :)

Because it's easier to think about, I'd prefer to represent it as a tree. (Also makes it easier to traverse the map and keep intermediate degrees.)
You could have a Node class, which would have a Collection of child Node objects. If you must, you could also represent the child relationships as a Relationship object, which would have both a weight and a Node pointer, and you could store a Collection of Relationship objects.
Then you could do a walk on the tree starting from the root, and mark each visited node with its degree.
class Node{
String name;
List<Relationship> children;
}
class Relationship{
Node child;
double weight;
}
class Tree{
Node root;
}
Here, Tree should probably have a method like this:
public Node findNodeByName(String name);
And Node should probably have a method like this:
public void addChild(Node n, double weight);
Then, as you parse each line, you call Tree.findNodeByName() to find the matching node (and create one if none exists... but that shouldn't happen, if your data is good), and append the subsequent items on the line to that node.
As you've pointed out, DAGs cannot really be converted to trees, especially because some nodes have multiple parents. What you can do is insert the same node as the child of multiple parents, perhaps using a hash table to decide if a particular node has been traversed or not.

Reading the comments, you seem confused by how a Node can contain Relationships which each in turn contains a Node. This is quite a common strategy, it is in general called the Composite pattern.
The idea in the case of trees is that the tree can be thought of as consisting of multiple subtrees - if you were to disconnect a node and all its ancestors from the tree, the disconnected nodes would still make a tree, though a smaller one. Thus, a natural way to represent a tree is to have each Node contain other Nodes as children. This approach lets you do many things recursively, which in the case of trees is often, again, natural.
Letting a Node keep track of its children and no other parts of the tree also emulates the mathematical directed graph - each vertex is "aware" only of its edges and nothing else.
Example recursive tree implementation
For instance, to search for an element in a binary search tree, you would call the root's search method. The root then checks whether the sought element is equal, less or greater than itself. If it is equal, the search exits with an appropriate return value. If it is less or greater, the root would instead call search on the left or right child, respectively, and they would do exactly the same thing.
Analogously, to add a new Node to the tree, you would call the root's add method with the new node as a parameter. The root decides whether it should adopt the new node or pass it on to one of its children. In the latter case, it would select a child and call its add method with the new Node as a parameter.

Efficient way to walk through an object tree?

I've got a set of TreeNodes, each of which has an id, a Collection of parent nodes, and a collection of child nodes.
For a given node Id, I'm looking for an efficient way to generate all the links that pass through that node. So in short, start at the node, and iterate through all its children. If a node has more than one child, create a link for each child. The traverse the children etc..
I'd also like to be able to do this in an 'upward' direction, through the parent nodes.
Is there a simple algorithm to do this?
EDIT: Oh, and I'd like to be able to output the id's of all the nodes in a given chain...

You are looking for a Breadth First or Depth First Search. At first it is not more than the following (this is depth first search).
Visit(Node node)
{
foreach (Node childNode in node.Children)
{
Visit(childNode);
}
DoStuff(node);
}
The problem is that the graph may contain cycles, hence the algorithm will enter infinite loops. To get around this you must remember visited nodes by flaging them or storing them in a collection. If the graph has no cycles - for example if it is a tree - this short algorithm will already work.
And by the way, if a TreeNode has multiple parents, it's not a tree but a graph node.

Well, if the nodes have a reference to the parent, it's simple as getting the parent recursively (once in a tree, each node has only one (or none at all, if it is a root) parent.
If there's no such reference, than you could use a breadth-first search, for instance, having as initial set your collection of parent nodes.
-- EDIT --
Once a node may have more than one parent, then you're dealing with a graph. There are also graph traversal algorithms (see table at the side).
Make sure that, if your graph has a cycle, you won't end up having a infinite loop

You might want to check out depthFirstEnumeration() and breadthFirstEnumeration() on DefaultMutableTreeNode. However, this doesn't solve your problem of wanting to navigate the tree in a bottom-up fashion.