I am going through B+ trees. I am having a doubt that what is the minimum number of keys in a B+ tree leaf node ?
I am reading from the following references, but they are saying something else
1). https://en.wikipedia.org/wiki/B%2B_tree
2). http://www.cburch.com/cs/340/reading/btree/
3). http://courses.cs.washington.edu/courses/cse326/08sp/lectures/11-b-trees.pdf
The last link uses some letter L for leaf node.
Can someone clarify, what must be the exact coount of keys in a leaf node ?
Min: 1, because a tree with just one entry must be possible.
Max: <branch factor> - 1, because that's how the branch factor is defined.
Related
I know what Binary search tree is and I know how they work. But what does it take for it to become a skewed tree? What I mean is, do all nodes have to go on one side? or is there any other combination?
Having a tree in this shape (see below) is the only way to make it a skewed tree? If not, what are other possible skewed trees?
Skewed tree example:
Also, I searched but couldn't find a good solid definition of a skewed tree. Does anyone have a good definition?
Figured out a skewed Tree is the worst case of a tree.
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The number of permutations of 1, 2, ... n = n!
The number of BST Shapes: (1/n+1)(2n!/n!n!)
The number of skewed trees of 1, 2, ....n = 2^(n-1)
`
Here is an example I was shown:
http://i61.tinypic.com/4gji9u.png
A good definition for a skew tree is a binary tree such that all the nodes except one have one and only one child. (The remaining node has no children.) Another good definition is a binary tree of n nodes such that its depth is n-1.
A binary tree, which is dominated solely by left child nodes or right child nodes, is called a skewed binary tree, more specifically left skewed binary tree, or right skewed binary tree.
I know well about Full Binary Tree and Complete Binary Tree. But unable to make Full binary tree with only 6 nodes.
The answer is No. You can't make a Full binary tree with just 6 nodes. As the definition in the Wikipedia says:
A full binary tree (sometimes referred to as a proper or plane
binary tree) is a tree in which every node has either 0 or 2
children. Another way of defining a full binary tree is a recursive
definition. A full binary tree is either:
A single vertex.
A tree whose root node has two subtrees, both of which are full binary trees.
Another interesting property I noticed is that, the number of nodes required to make a full binary tree will always be odd.
Another way to see that a full binary tree has an odd number of nodes:
Starting with the definition of a full binary tree (Wikipedia):
a tree in which every node has either 0 or 2 children.
This means that the total number of child nodes is even (0+2+2+0+...+2 is always even). There is only one node that is not a child of another, which is the root. So considering that node as well, the total becomes odd.
By consequence there is no full binary tree with 6 nodes.
Elaborating on #vivek_23's answer, this is, unfortunately, not possible. There's a beautiful theorem that says the following:
Theorem: Any full binary tree has 2L - 1 nodes, where L is the number of leaf nodes in the tree.
The intuition behind this theorem is actually pretty simple. Imagine you take a complete binary tree and delete all the internal nodes from it. You now have a forest of L single-node full binary trees, one for each leaf. Now, add the internal nodes back one at a time. Each time you do, you'll be taking two different trees in the forest and combining them into a single tree, which decreases the number of trees in the forest by one. This means that you have to have exactly L - 1 internal nodes, since if you had any fewer you wouldn't be able to join together all the trees in the forest, and if you had any more you'd run out of trees to combine.
The fact that there are 2L - 1 total nodes in a full binary tree means that the number of nodes in a full binary tree is always odd, so you can't create a full binary tree with 6 nodes. However, you can create a full binary tree with any number of odd nodes - can you figure out how to prove that?
Hope this helps!
Problem : Given a rooted Tree T containing N nodes. Each node is numbered form 1 to N, node 1 being the root node. Also, each node contains some value. We have to do three kind of queries in the given tree.
Query 1::
Given a node nd, you have to find the sum of the values of all the nodes of the subtree rooted at nd and print the answer.
Query 2::
Given a node nd, you have to delete the subtree rooted at nd, completely (including node nd).
Query 3::
Given a node nd and some integer v, you have to add a child to node nd having value equal to v.
Constraints : N will be of the order of 100000. And total number of queries wil also be of the order 100000. So, I can't to DFS traversal every time.
My Idea: My solution is offline . I will first find all the nodes that are added to the tree at-least once and make the corresponding tree. Then I will do pre-order traversal to the tree and convert it into an array where a subtree will always appear continuously. Then I can use segment tree data structure to solve the problem. My algorithm will be thus O(QlogN), where Q is the total number of queries. However, I am looking for a "online" solution which is efficient. I mean, I have perform each query as soon as it is asked. I can not store all the queries first then perform them one by one.
Any help is appreciated a lot!
Thanks.
Assuming tree is balanced, with two extra parameters in every node you can solve it in o(qlogn).
With every node maintain a sum whose value will be equal to the sum of values of nodes in the subtree rooted at that and maintain parent as well.
With the above two requirements, query one just reduces to returning sum plus the value at that node(o(1)). query two reduces to just subtracting sum plus the value of node from every parent of that node till you reach the root(o(logn)). query three just reduces to adding v to every parent of that node till you reach the root(o(logn)).
Suppose I am asked to construct a B+ tree, of:
i) n = x.
ii) order = x.
iii) degree = x.
iv) p = x.
What should the no. of keys, and pointers that each node can contain, in each of the above cases?
In B+ tree, Order denotes Maximum number of child pointers for each internal node, i.e. if Order of a B+ tree is m, then each internal node can have at most m children (subsequently, m-1 number of keys) and at least CEIL(m/2) number of children pointers (Except root).
For Degree of B+ tree, from this, I got the information that if d is the degree of a B-Tree, then each node can contain upto 2d items (keys). Now, both B tree and B+ tree are Multiway Tree, and hence, I suppose definition of degree will not change. Check the $LINK given as Comment also which indicates same fact.
For n, as JustinDanielson mentioned, it might be total number of keys stored in the node, for which number of children pointer would be n+1 (=x+1 for your question)
here is the Balancing Act problem that demands to find the node that has the minimum balance in a tree. Balance is defined as :
Deleting any node
from the tree yields a forest : a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created by deleting that node from T
For the sample tree like :
2 6 1 2 1 4 4 5 3 7 3 1
Explanation is :
Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The
larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node
1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees
has two nodes, so the balance of node 1 is two.
What kind of algorithm can you offer to this problem?
Thanks
I am going to assume that you have had a looong look at this problem: reading the solution does not help, you only get better at solving these problems by solving them yourself.
So one thing to observe is, the input is a tree. That means that each edge joins 2 smaller trees together. Removing an edge yields 2 disconnected trees (a forest of 2 trees).
So, if you calculate the size of the tree on one side of the edge, and then on the other, you should be able to look at a node's edges and ask "What is the size of the tree on the other side of this edge?"
You can calculate the sizes of trees using dynamic programming - your recurrence state is "What edge am I on? What side of the edge am I on?" and it calculates the size of the tree "hung" at that node. That is the crux of the problem.
Having that data, it is sufficient to iterate through all the nodes, look at their edges and ask "What is the size of the tree on the other side of this edge?" From there, you just pick the minimum.
Hope that helps.
You basically want to check 3 things for every node:
The size of its left subtree.
The size of its right subtree.
The size of the rest of the tree. (size of tree - left - right)
You can use this algorithm and expand it to any kind of tree (different number of subnodes).
Go over the tree in an in-order sequence.
Do this recursively:
Every time you just before you back up from a node to the "father" node, you need to add 1+size of node's total sub trees, to the "father" node.
Then store a value, let's call it maxTree, in the node that holds the maximum between all its subtrees, and the (sum of all subtrees)-(size of tree).
This way you can calculate all the subtree sizes in O(N).
While traversing the tree, you can hold a variable that hold the minimum value found so far.