What is the number of different binary trees and Binary search trees that can be formed from n nodes?
Pls Note :
1) I am asking for Binary trees and not full Binary trees (in which case the answer is Catalan(n))?
2) In case of BST's again pls include all the cases (including linear chains)
I think (expect) that Ans1 = Ans 2 * factorial (n), as each structure would only entail a single arrangement of keys as per BST ordering
Related
for the time efficiency of inserting into binary search tree,
I know that the best/average case of insertion is O(log n), where as the worst case is O(N).
What I'm wondering is if there is any way to ensure that we will always have best/average case when inserting besides implementing an AVL (Balanced BST)?
Thanks!
There is no guaranteed log n complexity without balancing a binary search tree. While searching/inserting/deleting, you have to navigate through the tree in order to position yourself at the right place and perform the operation. The key question is - what is the number of steps needed to get at the right position? If BST is balanced, you can expect on average 2^(i-1) nodes at the level i. This further means, if the tree has k levels (kis called the height of tree), the expected number of nodes in the tree is 1 + 2 + 4 + .. + 2^(k-1) = 2^k - 1 = n, which gives k = log n, and that is the average number of steps needed to navigate from the root to the leaf.
Having said that, there are various implementations of balanced BST. You mentioned AVL, the other very popular is red-black tree, which is used e.g. in C++ for implementing std::map or in Java for implementing TreeMap.
The worst case, O(n), can happen when you don't balance BST and your tree degenerates into a linked list. It is clear that in order to position at the end of the list (which is a worst case), you have to iterate through the whole list, and this requires n steps.
I have two questions:
What is the difference between nearly balanced BST and nearly Complete Binary tree. Even if the definition of the former is clear then we can differenciate, but not able to get a relevant article.
Today, in my class I was taught about the condition to be balanced as:
max( size(root.left) , size(root.right) ) <= 3*n/4 ------------ (eqn 1).
Hence, H(n) = height for the tree of n nodes following the above property < = 1+H(3*n/4).
Continuing the recursive steps we get the bound for logn.
My question is that, is this a specific type of BST ? For example in case of AVL trees, as I remember the condtion is that the difference in the heights of left and right childs being atmost 1, or is this a more general result and the equation 1 as stated earlier can be reduced to prove the result for AVL Trees as well ? i.e. any Balanced BST will result in difference of heights of siblings being atmost 1 ?
In case its different than AVL, how do we manage the Insertion and Delete Operations in this new kind of tree ?
EDIT : Also if you can explain why 3*n/4 only ?
My Thought: It is because we can then surely say that H(n) <= 1+H(3*n/4), since if we take something like 3n/5 less than 3n/4 then H(3n/5) wont be necessarily less that H(2n/5) as the raio of 3n/5 and 2n/5 is less than 2 and as we know a factor of 2 for number of nodes increases the height by 1.
So we wont surely write H(n) <= 1 + H(3n/5), it may be H(2n/5) in place of H(3n/5) as well, am I right ?
A nearly complete BST is a BST where all levels are filled, except the last one. Definitions are kind of messed up here (some call this property perfect). Please refer to wikipedia for this.
Being balanced is a less strict criterion, i.e. all (nearly) complete BSTs are balanced, but not all balanced BSTs are complete. In that Wikipedia article is a definition for that too. Im my world a BST is balanced, if it leads to O(log n) operation cost.
For example, one can say, a BST is balanced, if each subtree has at most epsilon * n nodes, where epsilon < 1 (for example epsilon = 3/4 or even epsilon = 0.999 -- which are practically not balanced at all).
The reason for that is that the height of such a BST is roughly log_{1/epsilon} n = log_2 n / (- log_2 epsilon) = O(log n), but 1 / (- log_2 0.99) = 99.5 is a huge constant. You can try to prove that with the usual ration of epsilon = 1/2, where both subtrees have roughly the same size.
I don't know of a common BST, which uses this 3/4. Common BSTs are for example Red-Black-Trees, Splay-Trees or - if you are on a hard disk - a whole family of B-Trees. For your example, you can probably implement the operations by augumenting each node with two integers representing the number of nodes in the left and the right subtree respectively. When inserting or deleting someting, you update the numbers as you walk from the root to the leaf (or up) and if the condition is validated, you do a rotation.
I have been working on determining the height of a self-balanced binary tree knowing its number of nodes(N) and I came with the formula:
height = ceilling[log2(N+1)], where ceilling[x] is the smallest integer not less than x.
The thing is I can't find this formula on the internet and it seems pretty accurate.
Is there any case of self-balanced binary tree this formula would fail?
What would be the general formula to determine the height of the tree then?
There is a formula on Wikipedia's Self-balancing binary search tree article.
h >= ceilling(log2(n+1) - 1) >= floor(log2(n))
The minimum height is floor(log2(n)). It's worth noting, however, that
the simplest algorithms for BST item insertion may yield a tree with height n in rather common situations. For example, when the items are inserted in sorted key order, the tree degenerates into a linked list with n nodes.
So, your formula is not far off from the "good approximation" formula (just off by 1), but there can be a pretty wide range between n and log2(n) to take into account.
Suppose I have to compare whether two binary search trees are similar. Now, the basic approach is the recursive formulation that checks for the root to be equal and then continues to check the equality of the corresponding right and left subtrees.
However, will it be correct to state that if the binary search trees have the same level order traversals then they are the same? Stated differently, does every BST have a unique level order traversal?
No, it isn't.
The first one:
1
\
\
2
\
\
3
The second:
1
/ \
/ \
2 3
Level order will give 1 - 2 - 3 for these two.
Since the informational theory lower bound on representing a binary tree with n nodes is 2n - THETA(log n), I don't think any simple traversal should be able to identify a binary tree.
Google search confirms the lower bound:
lower bound bits binary tree
There is a simple reduction from BST to binary tree. Consider the BSTs with nodes value 1..n. The number of these BSTs is the number of binary trees with n nodes (you could always do a pre order traversal and insert the value in that order). If you can use a level order traversal to identify such a BST, you can use 1 for a "in-level" node, 0 for a "end-level" node. The first tree becomes "000", the second one "010". This will let a BST be identified with just n bits, with does not fit the information theory lower bound.
Well , I discussed this question with a friend of mine , so the answer isn't exactly mine! , but here's what came up, the level order traversal you do for a BST can be sorted and thus you can get the inorder traversal of the particular BST. Now you get two traversals which can then be used to uniquely identify the BST. Thus it wouldn't be incorrect to state that every BST has a unique level order traversal.
Algorithm:
ConstructBST(levelorder[] , int Size)
1. Declare array A of size n.
2. Copy levelorder into A
3. Sort A
From two traversals A and levelorder of a Binary Search Tree , of which one is inorder, construct the tree.
We always see operations on a (binary search) tree has O(logn) worst case running time because of the tree height is logn. I wonder if we are told that an algorithm has running time as a function of logn, e.g m + nlogn, can we conclude it must involve an (augmented) tree?
EDIT:
Thanks to your comments, I now realize divide-conquer and binary tree are so similar visually/conceptually. I had never made a connection between the two. But I think of a case where O(logn) is not a divide-conquer algo which involves a tree which has no property of a BST/AVL/red-black tree.
That's the disjoint set data structure with Find/Union operations, whose running time is O(N + MlogN), with N being the # of elements and M the number of Find operations.
Please let me know if I'm missing sth, but I cannot see how divide-conquer comes into play here. I just see in this (disjoint set) case that it has a tree with no BST property and a running time being a function of logN. So my question is about why/why not I can make a generalization from this case.
What you have is exactly backwards. O(lg N) generally means some sort of divide and conquer algorithm, and one common way of implementing divide and conquer is a binary tree. While binary trees are a substantial subset of all divide-and-conquer algorithms, the are a subset anyway.
In some cases, you can transform other divide and conquer algorithms fairly directly into binary trees (e.g. comments on another answer have already made an attempt at claiming a binary search is similar). Just for another obvious example, however, a multiway tree (e.g. a B-tree, B+ tree or B* tree), while clearly a tree is just as clearly not a binary tree.
Again, if you want to badly enough, you can stretch the point that a multiway tree can be represented as sort of a warped version of a binary tree. If you want to, you can probably stretch all the exceptions to the point of saying that all of them are (at least something like) binary trees. At least to me, however, all that does is make "binary tree" synonymous with "divide and conquer". In other words, all you accomplish is warping the vocabulary and essentially obliterating a term that's both distinct and useful.
No, you can also binary search a sorted array (for instance). But don't take my word for it http://en.wikipedia.org/wiki/Binary_search_algorithm
As a counter example:
given array 'a' with length 'n'
y = 0
for x = 0 to log(length(a))
y = y + 1
return y
The run time is O(log(n)), but no tree here!
Answer is no. Binary search of a sorted array is O(log(n)).
Algorithms taking logarithmic time are commonly found in operations on binary trees.
Examples of O(logn):
Finding an item in a sorted array with a binary search or a balanced search tree.
Look up a value in a sorted input array by bisection.
As O(log(n)) is only an upper bound also all O(1) algorithms like function (a, b) return a+b; satisfy the condition.
But I have to agree all Theta(log(n)) algorithms kinda look like tree algorithms or at least can be abstracted to a tree.
Short Answer:
Just because an algorithm has log(n) as part of its analysis does not mean that a tree is involved. For example, the following is a very simple algorithm that is O(log(n)
for(int i = 1; i < n; i = i * 2)
print "hello";
As you can see, no tree was involved. John, also provides a good example on how binary search can be done on a sorted array. These both take O(log(n)) time, and there are of other code examples that could be created or referenced. So don't make assumptions based on the asymptotic time complexity, look at the code to know for sure.
More On Trees:
Just because an algorithm involves "trees" doesn't imply O(logn) either. You need to know the tree type and how the operation affects the tree.
Some Examples:
Example 1)
Inserting or searching the following unbalanced tree would be O(n).
Example 2)
Inserting or search the following balanced trees would both by O(log(n)).
Balanced Binary Tree:
Balanced Tree of Degree 3:
Additional Comments
If the trees you are using don't have a way to "balance" than there is a good chance that your operations will be O(n) time not O(logn). If you use trees that are self balancing, then inserts normally take more time, as the balancing of the trees normally occur during the insert phase.