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.
I have a red-black tree with n nodes, rooted at x. How can I prove or disprove that the number of nodes in the left subtree of x (including the root of x.left) is at most 3n / 4 without counting?
You can just construct a counter-example with as many red nodes as possible on the left, and no red nodes at all on the right.
If the right is a complete all black tree with 2^h-1 nodes, and the left can be a complete tree with 2^(2h)-1 nodes.
When h>=3 the left side has more than 3n/4 nodes.
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 have to find a family of maximally unbalanced red-black trees and to prove the "respective attributes" of that family to prove that there is an infinitely big family of red black trees that have a height close to 2log(n+1).
Now my guess is that this family consists of basically all the red black trees that have one path with s-r-s-r ... nodes and the rest filled with black nodes. But how do I prove this? and how do i formally write down how such a family looks like?
Thank you!
Now my guess is that this family consists of basically all the red black trees that have one path with s-r-s-r ... nodes and the rest filled with black nodes.
That's a reasonable guess.
But how do I prove this?
Describe an infinite sequence of trees T_0, T_1, T_2, T_3, ..., such that, for every integer n, there exists a tree in the sequence with at least n nodes. Show that there exists a constant C such that, for every i, the height of T_i is at least 2log(n_i+1) - C, where n_i is the number of nodes in T_i. (This is one possible interpretation of the ambiguous term "close to".)
how do i formally write down how such a family looks like?
Inductively. I'll do the all-black trees as an example. The tree T_0 is empty (base case). For all integers i > 0, the tree T_i consists of a black node with left and right subtrees equal to T_{i-1} (inductive step). Then you can prove facts about these trees using induction.