According to my textbook when N nodes are stored in a binary tree H(max) = N
According to outside sources when N nodes are stored in a binary tree H(max) = N - 1
Similarly
According to my textbook when N nodes are stored in a binary tree H(min) = [log2N+1]
According to outside sources when N nodes are stored in a binary tree H(min) = [log2(N+1)-1]
Which one is right and which is wrong? Are they supposed to be used in different situations. In this case what would be the maximum height of a tree with 32 nodes?
I have been looking through my resources to understand this concept, and for some reason all my sources have different answers. I can calculate height when a binary tree is pictorially represented because that would involve number of nodes in each subtree. What about when only the number of nodes are given?
Obviously it has to do with the definition of height. If the height is defined by the number of nodes traversed from top to bottom then the max height is N. If it's defined as the number of hops between nodes, then it's N-1. Same goes for minimum height. That said, what counts is it's respectively O(N) and O(log N).
Related
I have been asked this question in an interview and I'm curious to know what will be the correct explanation for it?
Consider the following height balanced BST where Balance factor = Height of left subtree - Height of right subtree & the accepted balance factors are 0, 1, -1, 2, and -2.
What will be the time taken to search an element in such a kind of height-balanced BST? Explain.
What I said was, even if it has a height factor of 2 rather than 1 in standard Balance BST definitions, still the operations should be logN complexity order, (where N is the number of elements in the tree) because when the N will be large then will be not making much difference if the height factor is 2 or 1.
If anyone can tell me what would have been the correct answer here will be helpful :)
We can solve this mathematically as :
Defining Worst Case
Now, in any Binary Tree, time complexity of searching is O(h) where h is height of the Binary Tree.
Now, for worst case, we want to find Maximum Height.
In case of simple Binary Search Tree with no Balancing Factor
Condition on Nodes, this maximum height can be n or n+1 (depending
on convention whether height of single node tree will be 1 or 0)
where n is number of nodes.
Thus, we can say that given number of nodes, worst case is maximum height.
Interestingly, we can also say that given height of a tree, the worst case is minimum nodes. As even for such minimum number of nodes, we might have to traverse down the tree of height h, which we also have to do for maximum number of nodes.
Thus, the intuition should be clear that Given Height of Tree, the worst case is minimum number of nodes.
Applying this Concept on Binary Search Tree
Let us try to construct Binary Search Tree of Height H such that number of nodes in the tree is minimum.
Here we will exploit the fact that Binary Tree is a Recursive Data
Structure (A Binary Tree can be defined in terms of Binary Tree)
We will use the notation NH to denote Minimum Number of Nodes in a Binary Search Tree of height H
We will create a Root Node
To Left (or Right) of Root, add a subtree of height H-1 (exploiting Recursive Property). So that number of nodes in entire tree is minimum, the number of node in Left (or Right) subtree should also be minimum. Thus NH is a function of
NH-1
Do we need to add anything to Right (or Left)?
No. Because there is no restriction of Balancing Factor on BST. Thus, our tree will look like
Thus, to construct Binary Search Tree of Height H such that number of nodes in the tree is minimum, we can take Binary Search Tree of Height H-1 such that number of nodes is
minimum, and can add 1 root node.
Thus, we can form Recurrence Relation as
NH = NH-1 + 1
with base condition as
N0=1
To create BST of height 0, we need to add one node. Throughout the answer we will use this convention
Now, this Recurrence Relation is quite simple to solve by Substitution and thus
NH = H+1
NH > H
Now, let n be the number of nodes in the BST of height H
Then,
n ≥ NH
n ≥ H
H ≤ n
Therefore,
H=O(n)
Or
O(H) = O(n)
Thus, Worst Case Time Complexity for Searching will be O(n)
Applying this Concept on AVL Tree
We can apply similar concept on AVL Tree. After reading later part of solution, one can find recurrence relation as :
NH = NH-1 + NH-2 + 1
with Base Condition :
N0 = 1
N1 = 2
And, inequality condition on solving recurrence will be
NH ≥ ((1+√5)/2)H
Then, let n be the number of nodes. Thus,
n ≥ NH
On simplifying, one can conclude that
H ≤ 1.44log2(n)
Applying this Concept on GIVEN Tree
Let us try to construct Given Tree of Height H such that number of nodes in the tree is minimum.
We will use the notation NH to denote Minimum Number of Nodes in Given Tree of height H
We will create a Root Node
To Left (or Right) of Root, add a subtree of height H-1 (exploiting Recursive Property). So that number of nodes in entire tree is minimum, the number of node in Left (or Right) subtree should also be minimum. Thus NH is a function of
NH-1
Do we need to add anything to Right (or Left)?
Yes! Because there is restriction of Balancing Factor on Nodes.
We need to add subtree on Right (or Left). What should be it's height?
H?
No, then height of entire tree will become H+1
H-1?
Permitted! since Balancing Factor of Root will be 0
H-2?
Permitted! since Balancing Factor of Root will be 1
H-3?
Permitted! since Balancing Factor of Root will be 2
H-4?
Not Permitted! since Balancing Factor of Root will become 3
We want minimum number of nodes, so out of H-1, H-2 and H-3, we will choose H-3. So that number of nodes in entire tree is minimum, the number of node in Right (or Left) subtree should also be minimum. Thus NH is also a function of
NH-3
Thus, to construct Given Tree of Height H such that number of nodes in the tree is minimum, we can have LEFT subtree as
Given Tree of Height H-1 such that number of nodes is minimum and can have RIGHT subtree as Given Tree of Height H-3 such that number of nodes in it is also minimum, and can add one Root Node. Our tree will look like
Thus, we can form Recurrence Relation as
NH = NH-1 + NH-3 + 1
with base condition as
N0=1
N1=2
N2=3
Now, this Recurrence Relation is Difficult to Solve. But courtesy to this answer, we can conclude that
NH > (√2)H
Now, let n be the number of nodes in the Given Tree Then,
n ≥ NH
n ≥ (√2)H
log√2(n) ≥ H
H ≤ log√2(n)
H ≤ 2log2(n)
Therefore, H=O(log(n))
Or O(H) = O(log(n))
Thus, Worst Case Time Complexity for Searching in this Given Tree will be O(log(n))
Hence, Proved Mathematically!
What is the Formula to Find the minimum number of vertices required to make a binary tree (not a complete binary tree) of height 5 ?
A binary tree's height cannot be bigger than the number of nodes or vertices in the tree. So yes, the minimum number of vertices required for a binary tree of height 5 will be 5. Also, there must be n-1 edges between them. You can imagine a single series of connected nodes, and that is basically what you get.
Alternately, a full binary tree is a binary tree in which each internal vertex has exactly two children.This means a binary tree with n internal vertices has 2n + 1 vertices, 2n edges, and n + 1 leaves.
the minimum number of vertices in a binary tree of height n (with no further constraints) is always n.
proof:
if there were less than n nodes, the tree wouldn't be a valid binary tree (I guess I don't need to further explain this point)
if there were more than n nodes, it means at least one of the nodes has a brother (pigeonhole principle), which could be taken off the tree and it would result in a valid, smaller binary tree, so we know that this tree is not minimal, which opposes our assumption of a minimal tree.
-> a binary tree T is minimal -> T has n nodes
We have a binary tree with n nodes. this tree is not necessarily balanced. for any node such x of this tree, we calculate the size (i.e: number of nodes) of left and right subtree of this node and set the label of this node as the minimum of these two values (values of right size and left size subtree). if any subtree has zero nodes, this size is equal to 0. Which of the following is True:
I) sum of labels belongs to order O(n log n).
II) there is a tree that sum of it's label belongs to order O(n). (i.e: is it possible to get a tree with sum of it's label be O(n)?)
III) there is a tree that sum of it's label belongs to order O(n^2).
My TA says two of these is true. My problem is with these sentences, anyone could describe it for me?
My guess is that it's 1 & 2.
1.) Consider the height of the tree. The height and the number of nodes n have the relationship n = (2^h)-1. From this relation, we can derive that h =logn. Now, Let's move to the number of nodes in each level of the binary tree. The maximum number of nodes that a level could have is (n/2) which is the last level (in a full binary tree, the last level will have n/2 nodes). So, the worst case of calculating the minimum is (number of levels)*(number of nodes in each level) => n/2logn => O(nlogn).
2.) It's possible to get an 0(n) solution by changing the height of the tree. For example of if considering a subset of the all nodes such that the height of the tree is zero/one, then it's possible to get an O(n) solution - for a tree with a total level of two, there can be a maximum of three nodes (root, left, right), and therefore, there's no minimum calculation involved in this. In this case, we don't have the additional logn in running time, and we end up with O(n).
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.
I am reading The Algorithm Design Manual. The author states that the height of a tree is:
h = log n,
where
h is height
n = number of leaf nodes
log is log to base d, where d is the maximum number of children allowed per node.
He then goes on to say that the height of a perfectly balanced binary search tree, would be:
h = log n
I wonder if n in this second statement denotes 'total number of leaf nodes' or 'total number of nodes'.
Which brings up a bigger question, is there a mathematical relationship between total number of nodes and the height of a perfectly balanced binary search tree?
sure, n = 2^h where h, n denote height of the tree and the number of its nodes, respectively.
proof sketch:
a perfectly balanced binary tree has
an actual branching factor of 2 at each inner node.
equal root path lengths for each leaf node.
about the leaf nodes in a perfectly balanced binary tree:
as the number of leafs is the number of nodes minus the number of nodes in a perfectly balanced binary tree with a height decremented by one, the number of leafs is half the number of all nodes (to be precise, half of n+1).
so h just varies by 1, which usually doesn't make any real difference in complexity considerations. that claim can be illustrated by remembering that it amounts to the same variations as defining the height of a single node tree as either 0 (standard) or 1 (unusual, but maybe handy in distinguishing it from an empty tree).
It doesn't really matter if you talk of all nodes or just leaf nodes: either is bound by above and below by the other multiplied by a constant factor. In a perfectly balanced binary tree the number of nodes on a full level is the number of all nodes in levels above plus one.
In a complete binary tree number of nodes (n) and height of tree (h) have a relationship like this in below.
n = 2^(h+1) -1
this is the all the nodes of the tree