According to this explanation of red black tree, the tree must have the following properties:
A node is either red or black.
The root is black. (This rule is sometimes omitted. Since the root can always be changed from red to black, but not necessarily
vice-versa, this rule has little effect on analysis.)
All leaves (NIL) are black. (All leaves are same color as the root.)
Both children of every red node are black.
Every simple path from a given node to any of its descendant leaves contains the same number of black nodes.
What is stopping someone making every single node black?
It is possible. But to maintain condition 5, sometimes you might want to color a node RED.
e.g., Consider the following example.
a
/ \
b c
Here all nodes can be BLACK
Now if you want to insert a new node, which color will you choose? RED. since if you choose black, the condition 5 will not be satisfied. So basically you can keep inserting RED nodes unless any of the conditions (1-4) is not broken
The last rule you quoted is "Every simple path from a given node to any of its descendant leaves contains the same number of black nodes."
If all nodes are black, then the path from the root to any leaf must contain the same number of nodes. In other words, all leaves are at the same depth - so this is only possible for a perfect binary 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'm working on data structures. There is something I do not understand about Red and Black trees. He always writes about the following features about these trees. But in all examples the values of the leaves are null. Why is that not in the features either. Why "All leaf nodes are black and blank." not?
Red/Black Property: Every node is colored, either red or black.
Root Property: The root is black.
Leaf Property: Every leaf (NIL) is black.
Red Property: If a red node has children then, the children are always black.
Depth Property: For each node, any simple path from this node to any of its descendant leaf has the same black-depth (the number of black nodes).
It doesn't really matter if all nodes are keyless and black or not.
If the nodes could be any color and/or empty, the asymptotics would not change at all, since red children cannot have red parents and all internal nodes have associated keys.
All paths would still have lengths that are at most a factor of two different in total number of keys, though now that would be 2L+2, rather than 2L for the longest path compared to the shortest path (of length L).
I am looking at the code of inserting into an augmented Red black tree. This tree has an additional field called "size" and it keeps the size of a subtree rooted at a node x. Here is the pseudocode for inserting a new node:
AugmentedRBT_Insert(T,x){
BST_Insert(T,x); //insert as if it is a normal BST
x[color]=red; //insert as a red node
size[x]=1;
tmp=parent[x];
while(tmp!=NULL){ //start from the node x and follow the path to root
size[tmp]=size[tmp]+1; //update the size of each node
tmp=parent[tmp];
}
}
Forget about fixing the coloring and rotations, they will be done in another function. My question is, why do we set the size of the newly added node "x" to 1? I understand that it will not have any subtrees, so its size must be 1, but one of the requirements of RBT is that every red node has two black children, in fact every leaf node is NULL and even if we insert the node "x" as black, it still should have 2 black NULL nodes and i think we must set its size to 3? Am i wrong?
Thanks.
An insertion in a red-black tree, as in most binary trees, happens directly at a leaf. Hence the size of the subtree rooted at the leaf is 1. The red node does have two black children, because leaves always have the "root" or "nil" as a child, which is black. Those null elements aren't nodes, so we wouldn't count them.
Then, we go and adjust the sizes of all parents up to the root (they each get +1 for the node we just added).
Finally, we fix these values when we rotate the tree to balance it, if necessary. In your implementation, you will probably want to do both the size updates and rotations in one pass instead of two.
I am taking an algorithms course and in my course slides, there is an example of insertion into a red-black tree:
My question is, why don't we let "2" be a leaf node here? It looks like if we let it be a leaf node, then no condition of a red black tree is violated. What am I missing here?
The Problem is not with the position of 2 the the second tree of your image but the color of different nodes. Here is the explanation:
1st Rule of insertion in Red-Black tree is: the newly inserted node has to be always Red. You fall in case 3 insertion where both the father and uncle of node 2 is Red. So they are needed to be recolored to Black, and the grandfather will become Red but as the grandfather is root so it will become Black again.
So the new tree (after inserting 2) should be like this (r and b indicate color, .b is Nil node):
5b
/ \
1b 7b
/ \ / \
.b 2r .b .b
/ \
.b .b
And why we always need to insert red node in RBT, you may ask? Answer is, 1st we know every NIL nodes are always Black in RBT, 2nd we have rule 5. Every simple path from a given node to any of its descendant leaves contains the same number of black nodes. Now if we insert a black node at the end the tree will violate this rule, just put 2b in above tree instead of 2r and keep color of 1 and 7 red, then count black node from root to any Nil node, you will see some path have 2 back nodes and some path have 3 black nodes.
All the leaves of a Red Black tree have to be NIL
Check property 3
The wikipedia article, based on the same idea, explains it as follow:
In many of the presentations of tree data structures, it is possible for a node to have only one child, and leaf nodes contain data. It is possible to present red–black trees in this paradigm, but it changes several of the properties and complicates the algorithms. For this reason, this article uses "null leaves",
So clearly nothing prevents you to do it your way, but you have to take it in account in your algorithms, which make them significantly more complex. Perhaps this issue can be somewhat alleviated by using OOP, where leaves contain elements, but behave as nodes with empty leaves.
Anyway, it's a trade off: what you would gain in space (roughly two pointers set to NULL in C), you'd probably lose in code complexity, computation time, or in the object runtime representation (specialized methods for the leaves).
Black-height not uniform.
If you count the number of blacks nodes searching NIL nodes from root, 5-1-2-nil has three and 5-7-nil or 5-1-nil only two.
(rule: Every path from a given node to any of its descendant NIL nodes contains the same number of black nodes)
I have an exam next week in algorithms, and was given questions to prepare for it. One of these questions has me stumped though.
"Can we draw a red-black tree with 7 black nodes and 10 red nodes? why?"
It sounds like it could be answered quickly, but I can't get my mind around it.
The CRLS gives us the maximum height of a RB tree with n internal nodes: 2*lg(n+1).
I think the problem could be solved using this lemma alone, but am not sure.
Any tips?
Since this is exam preparation, I don't want to give you a direct answer, but I think what you need to consider is the properties that govern how you build a Red-Black Tree:
A node is either red or black.
The root is black. (This rule is sometimes omitted from other definitions. Since the root can always be changed from red to black but not necessarily vice-versa this rule has little effect on analysis.)
All leaves are black.
Both children of every red node are black.
Every simple path from a given node to any of its descendant leaves contains the same number of black nodes.
(Stole these from the wikipedia page: http://en.wikipedia.org/wiki/Red-black_tree)
Given the count of nodes you listed, can you meet all of these properties?
the answer is simple.
As we know that a red node can have only black parent.The max no. of nodes will be when each black node's both children are red and, hence, every black node has red parent. So, for 'n' black nodes '2n' red node are possible.
Think it this way:
put the first node(which is root) & make it black
make both its children red
make left and right children of both these red nodes black and for all these black nodes,
follow the same procedure as followed with root until black node count reaches the given value (in this case 7)
hope this helped you visualize the solution.
The answer critically depends on whether your RB tree uses black dummy nodes at the leaves, and if so, they are included in the count of seven black nodes. If not, consider a complete tree of seven black nodes
*
/ \
* *
/\ /\
* * * *
You won't have much trouble adding ten red nodes.