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
