How Red Black Tree will formed if we insert same elements, for example inserting 8,8,8,8,8,8,8 .
If you insert 8,8,8,8,8,8,8 it will be displayed something like below,
because Red-Black Tree follows these properties:
A red-black tree is a balanced binary search tree with the following properties:
Every node is colored red or black.
Every leaf is a NIL node, and is colored black.
If a node is red,
then both its children are black.
Every simple path from a node to a
descendant leaf contains the same number of black nodes.
Related
I have nodes and edges in two different state objects ordered by when they are added in canvas in React-Flow-Renderer. now I want to get JSON of nodes sorted by their position (Top-Left to bottom right). How can I achieve this without effecting performance?
I'm trying to make a node rotate along the X axis to look at another node.
I tried using the SCNLookAtConstraint with the gimbal turned off, but this still allows the node to rotate on both the X and Y axes. (Also, it makes the rear of the node face the target, not the front.)
How do I calculate how to rotate one node to face another from two vector 3 positions?
The docs talk about the orientation of a node, and what it means to 'look' at another node:
A node points in the direction of the negative z-axis of its local coordinate system. This axis defines the view direction for nodes containing cameras and the lighting direction for nodes containing spotlights or directional lights, as well as the orientation of the node’s geometry and child nodes. When Scene Kit evaluates a look-at constraint, it updates the constrained node’s transform property so that the node’s negative z-axis points toward the constraint’s target node.
You can modify that by specifying a different value for constraint.localFront, such as SCNVector3(0,0,1) to point with the positive z-axis instead of the negative.
If you need more control over which axis you want to engage, then you have a couple of options.
Create an invisible target node that remains positioned along the plane perpendicular to your rotation axis
Instead of using constraints, use SCNNode.look(at:) to update the node within your game loop, providing a translated target coordinate along the perpendicular plane.
If the target is the camera, then checkout SCNBillboardConstraint, which allows you to specify which axes are free to rotate:
let billBoard = SCNBillboardConstraint()
billBoard.freeAxes = [.X]
node.constraints = [billBoard]
I'm making presentation for AVL tree, can't understand what is the balance factor.
please give me the link or any thing that I can understand graphically how height of an AVL tree's height effect
If I understood your question correctly.
Difference between the height of the left sub tree and right sub tree is the balance factor of an AVL tree.when the factor is 1,0, or -1 the tree is balanced otherwise unbalanced .
Play with AVL tree applet to get some intuition on this
See this link for Balance Factor
According to my book, coloring red links that go from nodes of even height to nodes of odd height in an AVL tree gives a (perfectly balanced) 2-3-4 tree, where red links are not necessarily left-leaning.
But I don't get how exactly. When I try it with some examples, I keep getting stuck. Can someone illustrate it for me please?
In my web application I work with ExtJS4 and I use an Ext.tree.Panel. For a custom drag and drop implementation I need to find the tree node by its screen coordinates.
Is there a way to find a tree node by its screen coordinate? Or is it possible to find it via its HTML element?
Thanx in advance...
It would be not very efficient to seek node by coordinates as the only way to find the node by coordinates is to walk over all nodes.
Finding node by HTML element seems like more logical approach to me. Extjs4 tree utilizes the view. So you can try the view.getRecord method:
var node = tree.view.getRecord(htmlEl);
I haven't tried this code but it should work provided you passed correct htmlEl.
One part of your question has already been answered.
If you really need to get HTML element (use it only as last resort) by screen coordinates, then you will need document.elementFromPoint.
Extra info and quirks here: http://www.quirksmode.org/blog/archives/2010/06/more_ie9_goodne.html