I just need confirmation that i´m right:
A binary tree with 'e' layers has
(2^e)-1
elements.
The other way around: A binary tree with 'n' elements has
log2(n+1)
layers...
Am i right?
It depends on how you describe the 'height' of your tree. Generally, a tree with only one element is said to be at height 0. The calculations are like this:
The number of leaves in a full binary tree of height n is 2^n. The number of interior nodes in a full binary tree of height n is 2^n - 1.
So a full binary tree with n layers will have a total of (2^n + 2^n - 1) elements. That turns out to be (2*2^n - 1) or simply 2^(n+1) - 1 elements.
The other way around, a binary tree with n elements has a height of log2(n+1) - 1.
Related
Given a binary tree the problem is to find all root-to-leaf paths. And we know the algorithm by passing the path in the form of a list and adding it to our result as soon as we hit a leaf.
My question How much space does storing all the path consumes. My intuition is that each path is going to consume memory order of the height of tree(O(h)), and if there are 2*n - 1 nodes in our full binary tree and then there are n leafs each corresponding to a path and so the space complexity would be O(n*log(n)) assuming the tree is height balanced. Is my analysis correct?
Your reasoning is correct, but it can be made more exact. A balanced binary tree is not necessarily a full binary tree.
Let N(h) be the number of paths when the height is h. Then N(h) ≤ 2 N(h - 1) This is because, given a tree of height h, the children are each trees of height at most h - 1. So
N(h) = O(2h).
Now we need to bound h. Since h appears in the exponent, it is not enough to find its order of growth. More exactly, it is known that
n ≥ 2h - 1
so
h ≤ log(n + 1)
Inserting this to what we have before
N(h) = O(2log(n + 1)) = O(n).
As you wrote the memory is the sum, per path, of the nodes on the path. The sum of nodes on each path is at most log(n + 1). Combining all the above gives O(n log(n)).
Actually, there is a direct result from the very definition of a tree: there is a unique path between any 2 nodes.
So, if n is the number of leaves, then total (root to leaf) paths = n.
And correspondingly, the tree's height is O(log n).
Is there a formula to calculate what the maximum and minimum height for an AVL tree, given a certain number of nodes?
For example:
Textbook question:
What is the maximum/minimum height for an AVL tree of 3 nodes, 5 nodes, and 7 nodes?
Textbook answer:
The maximum/minimum height for an AVL tree of 3 nodes is 2/2, for 5 nodes is 3/3, for 7 nodes is 4/3
I don't know if they figured it out by some magic formula, or if they draw out the AVL tree for each of the given heights and determined it that way.
The solution below is appropriate for working things out by hand and gaining an intuition, please see the exact formulas at the bottom of this answer for larger trees (54+ nodes).1
Well the minimum height2 is easy, just fill each level of the tree with nodes until you run out. That height is the minimum.
To find the maximum, do the same as for the minimum, but then go back one step (remove the last placed node) and see if adding that node to the opposite sub-tree (from where it just was) violates the AVL tree property. If it does, your max height is just your min height. Otherwise this new height (which should be min height+1) is your max height.
If you need an overview of what the properties of an AVL tree are, or just a general explanation of an AVL tree, Wikipedia is a great place to start.
Example:
Let's take the 7 node example case. You fill in all levels and find a completely filled tree of height 3. (1 at level 1, 2 at level 2, 4 at level 3. 1+2+4=7 nodes.) That means 3 is your minimum.
Now find the max. Remove that last node and place it on the left subtree instead of the right. The right subtree still has height 3, but the left subtree now has height 4. However these values differ by less than 2, so it is still an AVL tree. Therefore your max height is 4. (Which is min+1)
All three examples worked out below (note that the numbers correspond to order of placement, NOT value):
Formulas:
The technique shown above doesn't hold if you have a tree with a very large number nodes. In this case, one can use the following formulas to calculate the exact min/max height2.
Given n nodes3:
Minimum: ceil(log2(n+1))
Maximum: floor(1.44*log2(n+2)-.328)
If you're curious, the first time max-min>1 is when n=54.
1Thanks to Jamie S for bringing this failure at larger node counts to my attention.
2Technically, the height of a tree is the longest path length (in edges) between the root and any leaf node. However the OP's textbook uses a common alternate definition of height as the number of levels in a tree. For consistency with the OP and Wikipedia, we use that definition in this post as well.
3These formulas are from the Wikipedia AVL page, with constants plugged in. The original source is Sorting and searching by Donald E. Knuth (2nd Edition).
It's important to note the following defining characteristics of an AVL Tree.
AVL Tree Property
The nodes of an AVL tree abide by the BST property
AND The heights of the left and right sub-trees of any node differ by no more than 1.
Theorem: The AVL property is sufficient to maintain a worst case tree height of O(log N).
Note the following diagram.
- T1 is comprised of a T0 + 1 node, for a height of 1.
- T2 is comprised of T1 and a T0 + 1 node, giving a height of 2.
- T3 is comprised of a T2 for the left sub-tree and a T1 for the right
sub-tree + 1 node, for a height of 3.
- T4 is comprised of a T3 for the left sub-tree and a T2 for the right
sub-tree + 1 node, for a height of 4.
If you take the ceiling of O(log N), where N represents the number of nodes in an AVL tree, you get the height.
Example) T4 contains 12 nodes. [ceiling]O(log 12) = 4.
See the pattern developing here??
**The worst-case height is
Lets assume the number of nodes is n
Trying to find out the minimum height of an AVL tree would be the same as trying to make the tree complete i.e. fill all the possible nodes at each level and then move to the next level.
So at each level the number of eligible nodes increases by 2^(h-1) where h is the height of the tree.
So at h=1, nodes(1) = 2^(1-1) = 1 node
for h=2, nodes(2) = nodes(1)+2^(2-1) = 3 nodes
for h=3, nodes(3) = nodes(2)+2^(3-1) = 7 nodes
so just find the smallest h, for which nodes(h) is greater than the given number of nodes n.
Now for the problem of maximum height of an AVL tree:-
lets assume that the AVL tree is of height h, F(h) being the number of nodes in the AVL tree,
for its height to be maximum lets assume that its left subtree FL and right subtree FR have a difference in height of 1(as it satisfies the AVL property).
Now assuming FL is a tree with height h-1 and FR be a tree with height h-2.
now the number of nodes in
F(h)=F(h-1)+F(h-2)+1 (Eq 1)
Adding 1 on both sides :
F(h)+1=(F(h-1)+1)+ (F(h-2)+1) (Eq 2)
So we have reduced the maximum height problem to a Fibonacci sequence. And these trees F(h) are called Fibonacci Trees.
So, F(1)=1 and F(2)=2
so in order to get the maximum height just find the index of the the number in the fibonacci sequence which is less than or equal to n.
So applying (Eq 1)
F(3)= F(2) + F(1)+ 1=4, so if n is between 2 and 4 tree will have height 3.
F(4)= F(3)+ F(2)+ 1 = 7, similarly if n is between 4 and 7 tree will have height 4.
and so on.
http://lcm.csa.iisc.ernet.in/dsa/node112.html
It is roughly 1.44 * log n, where n is the number of nodes.
For a more detailed description on how that was derived. You can refer to this link starting on the middle of page 13: http://www.compsci.hunter.cuny.edu/~sweiss/course_materials/csci335/lecture_notes/chapter04.2.pdf
I have the following problem:
For a B-tree of degree 1, what is the maximum height for a the tree as
a function of the number of keys n in the tree?
And I thought that because the order of the tree is 1 that means the number of keys can be between 1 and 2. Therefore I took a tree with only 1 key in each node so I can have the maximum height. Adding the number of nodes for each level I got that
2^0+2^1+2^2+...+2^h= n, where n is the number of nodes and h is the height of the tree
and solving it I got that the height h is log(n+1) but I'm not really sure this is the right answer.
Any ideas?
Height of binary tree h=log(n+1)-1
Here is the derivation
Here i am assuming height of root is zero
so
n=2^0+2^1+2^2........2^h
Apply Geometric progression formula & get
h=log(n+1)-1.
Here log base is 2.
So when there is only single node at every level. We can get log(2)base 2, n times SO Maximum height becomes
h=n-1
The worst case height for a B-Tree of order m is logm/2n.
See: http://en.wikipedia.org/wiki/B-tree#Best_case_and_worst_case_heights
The binary-search algorithm takes log(n) time, because of the fact that the height of the tree (with n nodes) would be log(n).
How would you prove this?
Now here I am not giving mathematical proof. Try to understand the problem using log to the base 2. Log2 is the normal meaning of log in computer science.
First, understand it is binary logarithm (log2n) (logarithm to the base 2).
For example,
the binary logarithm of 1 is 0
the binary logarithm of 2 is 1
the binary logarithm of 3 is 1
the binary logarithm of 4 is 2
the binary logarithm of 5, 6, 7 is 2
the binary logarithm of 8-15 is 3
the binary logarithm of 16-31 is 4 and so on.
For each height the number of nodes in a fully balanced tree are
Height Nodes Log calculation
0 1 log21 = 0
1 3 log23 = 1
2 7 log27 = 2
3 15 log215 = 3
Consider a balanced tree with between 8 and 15 nodes (any number, let's say 10). It is always going to be height 3 because log2 of any number from 8 to 15 is 3.
In a balanced binary tree the size of the problem to be solved is halved with every iteration. Thus roughly log2n iterations are needed to obtain a problem of size 1.
I hope this helps.
Let's assume at first that the tree is complete - it has 2^N leaf nodes. We try to prove that you need N recursive steps for a binary search.
With each recursion step you cut the number of candidate leaf nodes exactly by half (because our tree is complete). This means that after N halving operations there is exactly one candidate node left.
As each recursion step in our binary search algorithm corresponds to exactly one height level the height is exactly N.
Generalization to all balanced binary trees: If the tree has less nodes than 2^N we for sure don't need more halvings. We might need less or the same amount but never more.
Assuming that we have a complete tree to work with, we can say that at depth k, there are 2k nodes. You can prove this using simple induction, based on the intuition that adding an extra level to the tree will increase the number of nodes in the entire tree by the number of nodes that were in the previous level times two.
The height k of the tree is log(N), where N is the number of nodes. This can be stated as
log2(N) = k,
and it is equivalent to
N = 2k
To understand this, here's an example:
16 = 24 => log2(16) = 4
The height of the tree and the number of nodes are related exponentially. Taking the log of the number of nodes just allows you to work backwards to find the height.
Just look up the rigorous proof in Knuth, Volume 3 - Searching and Sorting Algorithms ... He does it far more rigorously than anyone else I can think of.
http://en.wikipedia.org/wiki/The_Art_of_Computer_Programming
You can find it in any good Computer Science library and on the bookshelves of many (very) old geeks.
Why is the height of a balanced binary tree equal to ceil(log2N) for N nodes?
w = width of base (maximum number of leaves)
h = height of tree (maximum number of edges from root to leaf)
Divide w by 2 (h times) to get to 1, which counts the single root node at top.
N = w + w/2 + ... + 1
N = 2h + ... + 21 + 20
= (1-2h+1) / (1-2) = 2h+1-1
log2(N+1) = h+1
Check: if N=1, h=0. If h=1, N=3.
This formula is for if the bottom level is full. N will not always be so great, but would still have the same height, h. So we must take the log's ceiling.
Can anyone please tell me how you find the min/max height of B trees, 2-3-4 trees and binary search trees?
Thanks.
PS: This is not homework.
Minimal and Maximal height of a 2-4 tree
For maximal height of a 2-4 tree, we will be having one key per node, hence it will behave like a Binary Search Tree.
keys at level 0 = 1
keys at level 1 = 2
keys at level 2 = 4 and so on. . . .
Adding total number of keys at each level we get a GP on solving which we will get the maximal height of the tree.
Hence, height = log2(n+1) - 1
Solving it for a total of 10^6 keys we will get :
⇒ 1 * (2^0+ 2^1 + 2^2 + . .. . . . +2^h) = 10^6
⇒ 1*(2^(h+1) - 1) = 10^6
⇒ h = log2(10^6 + 1) - 1
⇒ Maximal height of 2-4 tree with a total of 10^6 keys is 19
For minimal height of a 2-4 tree, we will be having three keys(maximum possible number) per node.
keys at level 0 = 3
keys at level 1 = 3*(4)
keys at level 2 = 3*(4^2) and so on . . .
Hence, height = log4(n+1) - 1
Adding total number of keys at each level we will get a GP on solving which we will get the minimal height. Solving it for a total of 10^6 keys we get:
⇒ 3 * (4^0+ 4^1 + 4^2 + . .. . . . +4^h) = 10^6
⇒ (4^(h+1) - 1) = 10^6
⇒ h = log4(10^6 + 1) - 1
⇒ Minimal height of 2-4 tree is 9
Binary Search Tree
For maximal height we will have a continuous chain of length n(total number of nodes) hence giving us a height equal to n-1(as height starts from 0).
For minimal height we will have a perfectly balanced tree and as solved earlier we will have a height equal to log2(n+1)-1
If you want to know the length of the longest branch you have to traverse the whole tree keeping note of "the longest branch so far".
Start from root node and look for its children
if it is having a child node then
Select the left most child and store others in any one data structure
else
if the height of that node is maximum til now
set it as max
end if
end if
Loop through all nodes of tree and whatever you get at last is the maximum height
Similar you can do for minimum
Minimal height of a binary tree is O(log n), maximal is O(n), depending on how balanced it is.
Wikipedia has a lovely bit about B Tree Heights.
I'm not familiar with 2-3-4 trees, but according to wikipedia they have similar isometry to red-black and B trees, so the above link should educate you on that as well.
As for B trees, the min/max heights depend on the branching factor chosen for the implementation.
Binary trees have a maximum height of n when input is inserted in order, the minimum height is of course log_2(n) when the tree is perfectly balanced. When input is inserted in random order the average height is about 1.39 * log_2 n.
I am not too familiar with b trees but the minimum height is of course log_m(n) when perfectly balanced (m is the number of children per node). According to Wikipedia the maximum height is log_(m/2)(n).
2-3 trees have a maximum height of log_2(n) when the tree consists of only 2-nodes and the minimum height is about log_3(n) [~0.631 log_2(n)] when the tree consists of only 3-nodes.