So I was understanding Catalan Number Concept and try to implement it in code using topDown DP approach, so the recurrence relation is from what I have learned,
f(N) signifies total count of binary search trees possible by N number of nodes
f(N) = f(i-1).f(N-i)
where,
i = 1.....N (i signifies root node and traverse through all the nodes),
N is total number of nodes present in the binary tree,
f(i-1) signifies number of left subtrees possible when i is a root node,
f(N-i) signifies number of right subtrees possible when i is a root node.
there's a mathematical formulae by which we can find f(N) value,
2NcN/(N+1)
.
And, total count of binary trees possible by N number of nodes,
(factorial of N) * f(N)
My Doubt:
what's the difference between binary trees and binary search trees? I feel like both are defines same meaning but there's a difference that I'm not aware of so help me.
What's the difference between binary trees and binary search trees?
There is no difference in the shape of such trees, but binary search trees put a constraint on the values that its nodes have, which doesn't exist for (just) binary trees:
The value of a node must not be less than any of the values in its left subtree
The value of a node must not be greater than any of the values in its right subtree
Another way to express this constraint is:
The in-order traversal of the values in the tree must be non-decreasing.
Practically this means that when you have been given:
A shape of a binary tree with π nodes
A set of π unique values that are present in the tree
...then there is only one possibly way to associate those values with the tree's nodes for it to be a binary search tree. If it does not have to be a binary search tree, then there are π! ways to make that association.
Conclusions:
Given π, then πΆπ (the πth Catalan number) represents:
the number of binary trees with π nodes without values (i.e. number of shapes of binary trees)
the number of binary search trees with π distinct values
The number of binary trees with π distinct values is π!πΆπ
Related
let the number of nodes be 3.
If a,b,c.. are in order c>a>b then possible avl trees are:
n=1 gives 1,n=2 gives 2..(look image)
As we know for a BST it is 2n C n/ (n+1).Have anyone tried to deduce a formula that can find the number of avl trees when the number of nodes are given.
example question:what is the number of possible avl trees with 11 nodes?
I doubt that simple formula exists. But you can find number of possible AVL trees with dynamic programming, filling 2D table, where n is number of nodes, h is tree height, then sum all non-zero n-nodes entries:
F(n, h) = Sum[by all possible i]{F(i,h-1)*F(n-1-i,h-1)} +
Sum[by all possible j]{F(j,h-1)*F(n-1-j,h-2)} +
Sum[by all possible k]{F(k,h-2)*F(n-1-k,h-1)}
Explanation: we can make n-nodes h-height AVL tree, connecting root node with two valid trees of equal height (h-1), or with (h-1) and (h-2) trees, or with (h-2) and (h-1) trees.
I'm working on an algorithm problem. Given n, generate all structurally unique binary search trees that store values 1...n. The solution was to enumerate each number i in the sequence, and use the number as the root, the subsequence 1β¦(i-1) on its left side would lay on the left branch of the root, and similarly the right subsequence (i+1)β¦n lay on the right branch of the root. Then construct the subtree from the subsequence recursively. This approach ensures that the BST constructed are all unique, since they have unique roots.
Now my question is: what if the trees are not limited to binary search trees, if it can be any binary tree. How would the solution be? I'd still want to go over all the cases with root i, where i = 1, ... n. The left subtree doesn't have to be in the range of 1...(i-1), right subtree doesn't have to be in the range of (i+1)...n. But then how to arrange them then? Create an arbitrary subset of (i-1) elements and apply??
Suppose you were given the following problem: given n disks, arrange them in unique binary-tree shapes. Then, following your correct reasoning in the question, you could say the following: I'll number the disks 1, 2, 3, .., n; then I'll (recursively) build the trees whose root is at disk #1, then disk #2, etc.
So the tree rooted-digraphs you (correctly) found really have nothing to do with the content in the nodes, let alone the question of whether the contents satisfied the BST invariant. Given your question here,
If the question is how many rooted digraphs exist, then it's the same as before.
If the question is how many combinations of rooted digraphs + node contents there are, then you just enumerate the rooted digraphs as you've done, and, for each one, enumerate the permutations of 1, 2, ... n.
If this is the case, and you don't need to enumerate, but rather approximate the number of such trees, then note that this is n! multiplied by the Catalan Numbers.
Use your algorithm for BSTs. This generates the unique shapes of trees. The shapes are unique, because for each root n there are n-1 elements in the left subtree, the rest are on the right.
Then, for each shape, there are n! orderings of elements. This gives the results for arbitrary trees.
Can we sort 7 numbers in 10 comparisons?
Depth of a binary tree with n node is? log(n)+1 or something else
If every node in a binary tree has either 0 or 2 children then the height of the tree is log(n): is it true or false?
Inserting an element into a binary search tree of size n takes time proportional to ------?
A binary tree not balanced can have all children at the right (for example), so the maximum height of a tree with n nodes is n
Same as 2.
If the tree is not balanced the insertion is proportional to the number of nodes that you need to traverse to find the correct position. Potentially n nodes.
The definition of balanced in this question is
The number of nodes in its left subtree and the number of nodes in its
right subtree are almost equal, which means their difference is not
greater than one
if given a n as the number of nodes in total, how many are there such trees?
Also what if we replace the number of nodes with height? Given a height, how many height balanced trees are there?
Well the difference will be made only by the last level, hence you can just find how many nodes should be left for that one, and just consider all possible combinations. Having n nodes you know that the height should be floor(log(n)) hence the same tree at depth k = floor(log(n)) - 1 is fully balanced, hence you know that is needs (m = sum(i=0..k)2^i) nodes, hence n-m nodes are left for the last level. Some definition of a balanced binary tree force "all the nodes to be left aligned", in this case it is obvious that there can be only one possibility, without this constraint you have combinations of 2^floor(log(n)) chooses n-m, because you have to pick which of the 2^floor(log(n)) possible slots you will assign with nodes, forcing a total of n-m nodes to be assigned.
For the height story you consider a sum of combinations of 2^floor(log(n)) chooses i as i goes from 1 to 2^floor(log(n)). You consider all possibilities of having either 1 node at the last level, then 2 and so on, until you don't make it a fully balanced binary tree, hence having all 2^floor(log(n)) slots assigned.
I know one cannot construct a tree without having both Inorder and Preorder/postorder traversals. Because for a given (only Inorder/Preorder/postorder) there could be a possibility of generating more number of trees. Are there any algorithms or mechanism one can compute the number of unique trees from a given (only Inorder/Preorder/postorder traversal).
Eg : a b c d e f g this is my Inorder traversal.
How many unique trees that can be constructed with the given Inorder traversal.
I tried them is google but none of the explanations are clear
Any help would be appreciated...
Well the algorithm is as follows:
Let, P(N) denote the number of trees possible with N nodes. Let the indexes of the nodes be 1,2,3,...
Now, lets pick the root of the tree. Any of the given N nodes can be the root. Say node i has been picked as root. Then, all the elements to the left of i in the inorder sequence must be in the left sub-tree. Similarly, to the right.
So, total possibilities are: P(i-1)*P(N-i)
In the above expression i varies from 1 to N.
Hence we have,
P(N) = P(0)*P(N-1) + P(1)*P(N-2) + P(2)*P(N-3)....
The base cases will be:
P(0) = 1
P(1) = 1
Thus this can be solved by using Dynamic Programming.
Note that a particular traversal is just a way of labeling the nodes in a tree, so that the number of possible binary trees is the same for any two traversals of the same length. The number of binary trees with n nodes is given by the n-1st Catalan number.
The formula
(2n)!/ (n)!(n+1)!
OR
2n * C(n) / (n+1)
gives the number of possible binary trees for any given INORDER/PREORDER/POSTORDER traversal.