I have some code where the user can delete one child record of a parent record one at a time. I'm detecting when there are no children left. At that point, I'm deleting the parent record as well. When coming up with a name for the variable, I realized I don't know what you call a parent with no children.
Is there a single, accepted name that I haven't heard of (or can't remember)?
Leaf: (no children)
Terminal: (never children)
Vacant: (had children)
Orphan: (no parent)
Leaf seems pretty standard according to wikipedia. "Terminal Node" on wikipedia will re-direct you to leaf node.
Vacant node in more detail:
Your orphanage had a child. The child got adopted and left. Your orpahage is now vacant.
Terminal node in more detail:
Example in OpenGL:
An EBO references a VAO.
A VAO references a VBO.
A VBO is raw data and references nothing else.
( EBO --> VAO --> VBO )
Thinking of the VBO as a "terminal node" would make more sense than thinking of it
as a leaf. Because it will not and cannot have children.
There are two common computer science notions of parent: relative/transitive and absolute/non-transitive. One node can be the parent of another (a child). (Hence, object of a transitive "is" or "parents"). A node with no children is a leaf node; a node with children is a parent node. (Hence, object of a non-transitive "is"). Neither of these involve the non-instantaneous notion of ever once having had children, eg as in the real world.
Each node of the type you are describing is during processing (before and after pruning) both a parent and not a parent. So you are using something like the non-instantaneous/real-world third meaning, something like "was a parent but isn't any more". (Think about what you're saying: it's a parent in the sense that it used to be a parent node even though it's not a parent node. So it's a parent ...because... it's not a parent.)
Once it's going to be deleted, how about NoLongerParent? But if you're just using a name for that node for all the processing, and it starts as a non-leaf, why not just Parent?
Related
From this post and this codebase, I know that there are pointers for
Youngest child
Youngest sibling
Oldest sibling.
So with Oldest child, how do I get?
I am thinking of access "children" pointer (current->children) and traverse to the end of that doubly linked list.
Get the oldest sibling of the youngest child:
current->p_cptr->p_osptr
I've been looking for the difference of ExitRule and EnterRule in Listener in "the definitive antrl4 reference" book but I still don't understand the difference. What is the difference between these two? And how does Listener travel the tree?
simply said, these are auto generated events created by Antrl
to keep track of the walker, so to speak.
Imagine, you stand in the middle of a long floor with countless doors on either site. Which you have to open and something in the rooms.
To keep track on which doors you've already visited, you mark them with a X and an incrementing number behind.
enterRule == You opens the door of a room.
you get in and search
exitRule == leave the room and paint the X and the next number on the door.
Now you are able to tell exactly which rooms you have already visited, but further more you are able to go to a specific room again for another search, without having to go all the way back and start all over again.
more technically spoken.
Antrl creates an enter and exit method for each Rule that is defined.
These Methods, or also known as callbacks, being used by the walker to walk the given tree.
Using a *ParseTreeListener you provide an entry point which indicates the beginning of the tree. For example enterAssign.
The Walker looks for this event and triggers it.
It then looks for a sub enterRule to trigger this event, and so on...
It keeps walking for as long as it find further enter rules or triggers exitAssign and the walker stops its walk.
Keypoint here is the automated or indipendent walk behavior.
The ParseTreeVisitor on the other hand,
will not generating enter/exit Rules to walk a tree.
It will generate visitRule instead.
The visit methods have to be called explicitly!
That means, if you forget to invoke a visit all its children don't get visited at all.
Antrl-Doc --> Parse Tree Listeners
Antrl-Mega-Tutorial --> many Information, short and precise
Tree Walking
Walking starts from a Root-Node and goes down on it until it have found the very left-most nested item. Then goes back up until it reaches the first node and looks for a sub-tree on the right.
It enters the right tree if one is found.
Or, it gets futher up the chain to find the next node.
...
Until it reaches the Root-Node again.
short Picture:
parent
|
/ \
/ \
Child1 Child2
/
/
Grandchild
chain of Calls:
enter parent
enter Child1
enter Grandchild
exit Grandchild
exit Child1
enter Child2
exit Child2
exit parent
React newbie again.
I have and admin page for a forum. It's a tree structure with potentially infinite depth, ordered by order_id field.
I can get it from the api either as a tree (with a children array field holding the children of each node) or as a flat array with an ancestry (a string that holds the whole path to the node, separated by a /) and a parent_id field.
I need to be able to move them around, expand and collapse subtrees and be able to add a node in any place.
Now I have several questions:
Is there a good out-of-the-box UI library to be able to drag-drop them to rearrange? I found this one: https://github.com/frontend-collective/react-sortable-tree but I can't get it to not have a limited height frame, and also customizing those themes looks too complicated.
If I am to do it from scratch using react-dnd, what is the more efficient way of working with this tree - have it as a tree (how do I add/update the nodes?) or store it flat and build the tree on render (easy to manage the collection, but complicates rendering and sorting logic).
This sounds like a complicated and interesting problem. If the library cannot do what you want it to stylistically then you will have to build it from scratch.
I would store all of your node objects in an Array with a children property that holds a reference to all of it's children; you could use their order_id as a reference if it is guaranteed to be unique.
I would also store the parent node's order_id as well, as it will help when you manipulate the tree.
Nodes = [
{order_id: 1, name:"order1", children_ids:[2,3], parent_id: 0},
{order_id: 2, name:"order2", children_ids:[], parent_id: 1},
{order_id: 3, name:"order3", children_ids:[], parent_id: 1},
]
You will have to build the tree on render, it may be a brain twister but it's more than doable with a bit of work.
When you add a new node, you will have to update the parent and children of it's new spot in the tree.
When you rearrange a node, you will have to travel up to the parent and remove it from it's children. You will also have to travel down to it's children and set their parents to the rearranged node's old parent. This removes the node from the tree. Then, add the rearranged node to the tree.
Any manipulation you do will involve manipulating the parent and children of that node - this is why I recommend storing both children's id and the parent's id. It saves you from having to do too many tree traversals which is not computationally efficient.
I don't recommend using an ancestry field as you mentioned. The logic for moving nodes around will be too complicated and you would have to traverse and edit potentially the entire tree if you do it this way.
The first step is to integrate with react-dnd; make each node draggable. When you drop it onto another node to rearrange, you trigger your rearrange handler.
Best of luck.
I understand the logic of using a queue to visit the nodes in a binary search tree level by level.
However i tried to implemment in C but i'am stuck because i don't know how to enqueue them properly. Starting with the root i can create a Queue but after that if i add the children of the root to the queue i will lose the children of those new nodes since iam modifying the connections in the Queue every time a add a new node.
I could create a new data type that has one more link to use in the linked list Queue, that should work. What is the best approach here?
visit[ing] the nodes in a binary search tree level by level
has a name: it is called a "breadth-first" traversal of the tree. Starting with an empty queue, you enqueue the root node and then repeatedly dequeue the first node in the queue, process it somehow, and enqueue all that node's children, until there are no more nodes enqueued. When exactly you should enqueue a node's children relative to other processing of that node may depend on exactly what processing you intend to perform, especially if it involves structurally modifying the tree.
As long as the per-node processing can affect only the subtree rooted at the then-current node, this is all fine. If you need to be able to affect other parts of the overall tree, however, then a breadth-first traversal probably is not appropriate for your task.
You said
[I] don't know how to enqueue them properly. Starting with the root [I] can create a Queue but after that if [I] add the children of the root to the queue [I] will lose the children of those new nodes since [I] am modifying the connections in the Queue every time a add a new node.
The key concept here is that membership and position in the queue are separate and independent from membership and position in the tree. You could manage that by adding additional links to the node structures themselves, or by creating a new structure for the queue elements that contains a pointer to the enqueued BST node. The latter decouples the tree from the queue, which many, including me, would consider preferable for most purposes.
My question is : how do you read the version of a (child) frozen node?
More details :
- Lets imagine the following situation : node A has two children node B and node C.
- All three nodes are versionable nodes (have the mixin type)
- Let's say node A has 7 versions from 1.0 to 1.6 : at version 1.2 we added node B, at version 1.4 we added node C.
I have a routine that prints the "output" of node A at version 1.5.
//I already obtained the Version object version associated to node A version 1.5
NodeIterator nodeIterator = version.getNodes();
while (nodeIterator.hasNext()) {
Node currentNode = nodeIterator.nextNode();
System.out.println("Node: "+currentNode.getPath());
// I can see that some of the nodes correspond to children (versions) of
//node B and C ..how can I get their version?
}
Thank you.
Each version in the history of the parent will contain a "frozen node" that represents a snapshot of the parent as it existed when the version was created (e.g., when the parent node was checked in). This frozen node contains child nodes that are either snapshots of the child (if the child was not versionable) or a "nt:versionedChild" node (if the child was versionable). This "nt:versionedChild" node contains a single "jcr:childVersionHistory" REFERENCE property that points to the version history of the child.
Section 3.13.10 of the JSR-283 specification contains a pretty good diagram that shows this structure.
Note that the "nt:versionedChild" node does not point to the particular version in the child's version history, which means that you have to determine which version of the child existed at the time the parent was checked in. After all, in this case of a versionable child, the parent and child version histories are independent and the nodes are checked in independently. (The reason might also stem from the fact that the application can change both the parent and child nodes in a single session, and then either check the parent in followed by checking the child in, or checking the child in followed by checking the parent in.)
One way to do this is by looking at the creation date of each version in the child's version history. This isn't ideal, as it will only tell you which child version existed at the time the parent was created. This could be ambiguous, since the comparison of timestamps depends on whether the parent was checked in before or after the versioned child. Within your application, you might have conventions that help you get around these ambiguities.
Perhaps a better option is to use version labels and apply the same label to the parent version and child version. (This is probably easier to do just after checking in the nodes rather than at a later time.) Then once you have a particular version of the parent, obtain its label and use it to find the corresponding version of the child in the child's version history (using VersionHistory.getVersionByLabel(String)).