Hi I know little about red-black tree before and maybe it's a stupid question.
Here's a 4-node tree:
Is it a legal red-balck tree?
In my opinion, it actually violates rules of red-black tree:
Every red node has both of its children colored black.
Every path from the root to a tree leaf contains the same number (the "black-height") of black nodes.
If not what should a 4-node red-black look like?
Thanks
You are correct that this tree violates the rules of red-black trees because the path 3-2-1-Null goes through 3 black nodes, while 3-4-Null only goes through 2.
Recall that there is no constraint that black nodes have to have their children painted red, only the reverse. Even a tree with all black nodes is technically a red-black tree, as long as it is balanced and therefore satisfies that "Every path from the root to a tree leaf contains the same number (the 'black-height') of black nodes."
As a result, you could make this tree a valid red-black tree by painting nodes 2 and 4 black and painting node 1 red. Notice node 1 (the only red node) would still have 2 black children and all paths from the root to leaves would hit 3 black nodes. Therefore satisfying the rules of a red-black tree.
Related
Properties of Red-Black Tree:
Every node is either red or black.
The root is black.
Every leaf (NIL) is black.
If a node is red, then both its children are black.
For each node, all simple paths from the node to descendant leaves contain the same number of black nodes.
According to the properties, are these valid or invalid red black trees?
A.
I think this is valid
B.
I think this is valid, but I am not sure since there two adjacent red nodes?
C.
I think this is valid, but I am not sure since there two adjacent red nodes?
D.
I think this is not valid since it violate Property 4?
Did I understand these properties of a RBtree right? If not, where am I wrong?
You have listed the properties of Red-Black trees correctly. Of the four trees only C is not a valid red-black tree:
A.
This is a valid tree. Wikipedia confirms:
every perfect binary tree that consists only of black nodes is a red–black tree.
B.
I think this is valid, but I am not sure since there two adjacent red nodes?
It is valid. There is no problem with red nodes being siblings. They just should not be in a parent-child relationship.
C.
I think this is valid, but I am not sure since there two adjacent red nodes?
It is not valid. Not because of the adjacent red nodes, but because of property 5. The node with label 12 has paths to its leaves with varying number of black nodes. Same for the node 25.
As a general rule, a red node can never have exactly one NIL-leaf as child. Its children should either both be NIL-leaves, or both be (black) internal nodes. This follows from the properties.
D.
I think this is not valid since it violate Property 4?
Property 4 is not violated: the children of the red nodes are NIL leaves (not visualised here), which are black. The fact that these red nodes have black NIL leaves as siblings is irrelevant: there are no rules that concern siblings. So this is valid.
For an example that combines characteristics of tree C and D, see this valid tree depicted in the Wikipedia article, which also depicts the NIL leaves:
A, B & D are valid red-black trees
C is not valid red-black tree as the black height from root to leaf is not the same. It is 2 in some paths and 1 in other paths. It violates what you stated as rule 5.
If 12 had a right child that was black and 25 a left child that was black, then it would be a red-black tree.
A red-black tree is basically identical to a 2-3-4 tree(4-Btree), even though the splitting/swapping method is upside down.
2-3-4 trees have fixed-size 3-node buckets. The color black means that it's the central node of the 3-bucket. Any red-black tree is considered as a perfect quadtree/binary tree (of 3-node-buckets) with empty nodes(black holes and red holes).
In other words, every black node (every 3-bucket) has its absolute position in the perfect tree(2 dimensional unique Cartesian or 4-adic/2-adic unique fraction number).
NIL nodes are just extra flags to save space; you don't have enough memory to store a perfect quadtree/binary tree.
The easiest way to check a red-black tree is to check that each black node is a new bucket(going down) and each red node is grouped with the above black node(same bucket). If the central black node has less than 2 red nodes, you can just add empty red holes next to the central black node(left and right).
A new black node is always the grandson of the last black node, and each black node can have only two red daughter-nodes and no black son-nodes. If the red daughter(mother) is empty(dead/unborn), the motherless grandson-node is directly linked to its grandfather-node.
A motherless black grandson-node has no brother, but he can have a black cousin-node next to him; the 2 cousins are linked to the same grandfather.
A quadtree is a subset of a binary tree.
All black nodes have even heights(2,4,6...), and all red nodes have odd heights(1,3,5...). Optionally, you can use the half unit 0.5.
The 3-bucket has a fixed size 3; just add extra red holes(unborn unlinked red daughters) to make the size 3.
I know there are two cases which uncle is black in red black trees when inserting a new node. But in all the cases parent is red. if parent is black there is no violation. what shall I do in such situation in a red black tree?
If you end up adding a node into a red/black tree and its parent is black, you can just make the node red and call it a day. There’s no need for any fixup. If you look at the rules for red/black trees, this doesn’t cause any new violations because all root-null paths still go through the same number of black nodes.
A different perspective: if you think of red/black trees as an isometry or 2-3-4 trees, then this rule corresponds to taking a leaf with one key in it and adding in another key, which doesn’t require any further fixup steps.
I am studying red black trees from CLRS.
I have 2 questions about the part where properties of red-black trees are discussed.
The passage from CLRS is as follows:
A red-black tree is a binary tree that satisfies the following red-black properties:
Every node is either red or black
The root is black
Every leaf(NIL) is black
If a node is red, then both its children are black
For each node, all simple paths from the node to descendant leaves contain the same number of black nodes
First of all, it says a red-black tree is a binary tree. Why didn't they say a red-black tree is a binary search tree. I thought the whole purpose of a red-black tree is to maintain the balance in a search tree to insure logN operations. Second, why is the leaf in a red-black tree the NIL?
In regular binary trees, we're used to this:
4
5 7
3 2 Nil Nil
Nil Nil Nil Nil
In this case, 3, 2, and 7 are the leaves. Why are the leaves depicted as the Nil's in CLRS?
In the book, they refer to a red-black tree as a binary search tree, it's right at the beginning of the chapter.
Also in the chapter, they indicate that nodes labeled as nilare actual nodes (called sentinels), with the same fields as an ordinary node but with a color field with fixed value "black" (the other fields can be set to arbitrary values); these nodes always appear as leafs in the tree. So it's different from the usual BST where a leaf is a node whose left subtree and right subtree are both nil.
The smallest number of internal nodes in a red-black tree with black height of k is 2k-1 which is one in the following image:
The largest number of internal nodes with black height of k is 22k-1 which, if the black height is 2, should be 24 - 1 = 15. However, consider this image:
The number of internal nodes is 7. What am I doing wrong?
(I've completely rewritten this answer because, as the commenters noted, it was initially incorrect.)
I think it might help to think about this problem by using the isometry between red-black trees and 2-3-4 trees. Specifically, a red-black tree with black height h corresponds to a 2-3-4 tree with height h, where each red node corresponds to a key in a multi-key node.
This connection makes it easier for us to make a few neat observations. First, any 2-3-4 tree node in the bottom layer corresponds to a black node with either no red children, one red child, or two red children. These are the only nodes that can be leaf nodes in the red-black tree. If we wanted to maximize the number of total nodes in the tree, we'd want to make the 2-3-4 tree have nothing but 4-nodes, which (under the isometry) maps to a red/black tree where every black node has two red children. An interesting effect of this is that it makes the tree layer colors alternate between black and red, with the top layer (containing the root) being black.
Essentially, this boils down to counting the number of internal nodes in a complete binary tree of height 2h - 1 (2h layers alternating between black and red). This is equal to the number of nodes in a complete binary tree of height 2h - 2 (since if you pull off all the leaves, you're left with a complete tree of height one less than what you started with). This works out to 22h - 1 - 1, which differs from the number that you were given (which I'm now convinced is incorrect) but matches the number that you're getting.
You need to count the black NIL leafs in the tree if not this formula won't work. The root must not be RED that is in violation of one of the properties of a Red-Black tree.
The problem is you misunderstood the black height.
The black height of a node in a red-black tree is the the number of black nodes from the current node to a leaf not counting the current node. (This will be the same value in every route).
So if you just add two black leafs to every red node you will get a red-black tree with a black height of 2 and 15 internal nodes.
(Also in a red-black tree every red node has two black children so red nodes can't be leafs.)
After reading the discussion above,so if I add the root with red attribute, the second node I add will be a red again which would be a red violation, and after node restructuring, I assume that we again reach root black and child red ! with which we might not get (2^2k)-1 max internal nodes.
Am I missing something here , started working on rbt just recently ...
It seems you havent considered the "Black Leaves" (Black nodes) -- the 2 NIL nodes for each of the Red Nodes on the last level. If you consider the NIL nodes as leaves, the Red nodes on the last level now get counted as internal nodes totaling to 15.
The tree given here actually has 15 internal nodes. The NIL black children of red nodes in last layer are missing which are actually called external nodes ( node without a key ). The tree has black-height of 2. The actual expression for maximum number of internal nodes for a tree with black-height k is 4^(k)-1. In this case, it turns out to be 15.
In red-black trees, external nodes[null nodes] are always black but in your question for the second tree you have not mentioned external nodes and hence you are getting your count as 7 but if u mention external nodes[null nodes] and then count internal nodes you can see that it turns out to be 15.
Not sure that i understand the question.
For any binary tree where all layers (except maybe last one) have max number of items we will have 2^(k-1)-1 internal nodes, where k is number of layers. At second picture you have 4 layers, so number of internal nodes is 2^(4-1)-1=7
I am searching for a proof that all AVL trees can be colored like a red-black tree?
Can anyone give the proof?
By definition R/B trees can be slightly less balanced then AVL-s, as |maxPath - minPath| must be <= 1 for AVLs and maxPath <= 2 * minPath for R/Bs so that not every R/B is an AVL but on the other hand there is no need for the AVL-s To have Empty subTrees so
4
/ \
3 6
/\ /\
1 E 5 8
is a perfectly legal AVL and it is not an R/B because R/B cannot contain Leaves and must be terminated by Empty trees which are coloured always Black - so that you cannot colour the tree above.
To make it R/B you are allowed to convert every leaf x into node E x E
and then follow these rules:
R/B Tree:
Must be a BST
must contain only nodes and empty trees which are coloured either Black or Red
Every Red node has black children
All Empty Trees are Black
Given a node, all paths to Empty Trees must have same number of Black Nodes
Any Leaf can be replaced with Node whose Left & Right subTrees are Empty
Max Path T ≤ 2 * Min Path T
Btw just realized it coloured my nodes and leaves in Red - this was not intended.
Karol
The red-black tree adjusts the branch heights by red nodes, so if the height difference is bounded, you can always adjust all the branches, starting from the shortest black branch. But it requires a very large cost, because you have to count all the branch heights.
The maximum local height difffernce bound of red-black tree is 2.
I suspect the answer is no.
AVL trees balance better than RB trees, which means they balance differently, which would rather imply that you could not colour every AVL tree as a valid RB tree.
The answer is yes, every AVL tree can be colored Red-Black, and the converse doesn't hold.
I haven'y exactly figured out HOW to do it tho, and am also seeking the proof.