In section 2 of http://tmc.web.engr.illinois.edu/heap_ianfest.pdf, Chan writes that for some alpha in (1/2, 1), the number of nodes at height (i + 1), n_{i+1} to be at most alpha * n_i, the number of nodes at height i.
If I start with the complete binary tree consisting of three nodes and an edge graph, the invariant holds (2 nodes at height 2 and 3 nodes at height 1). But inserting a node will result in 4 nodes at height 1 and the invariant no longer holds. Is this right?
Your understanding of this is correct. The invariant applies specifically after a call to delete-min() when any non-conforming trees are broken up to restore the invariant.
The trick to reconciling this with the rest of the analysis is noting where the invariant is required for the data structure to function.
The invariant is responsible for showing that the maximum height of a tree is no more than log_{1/alpha}(n). Note that if you have a configuration of the data structure that conforms to the invariant, such as your 3 node edge graph example, and you insert() another node (or alternatively, call decrease-key()), there cannot be any tree in the resulting configuration that has a height greater than log_{1/alpha}(n).
The other place that the invariant is used is in the amortized analysis for delete-min(), after which the invariant by definition holds.
Related
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).
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!
The fan-out of a node in a tree is defined to be the number of children the node has. The fan-out of a tree is defined to be the maximum fan-out of any node in the tree. Assume tree T has n nodes and a fan-out f > 1. What is the minimum possible height of T?
I have no idea how to begin this problem. I solved the first part which is to find the maximum number of nodes that can be T in terms of height h and fan-out f > 1. I got (f^(h+1) -1)/(f-1). I'm thinking you can use this to solve for the question above. Can someone please point me in the correct direction?
Thank you!
I would approach this problem by turning it around and trying to find the maximum number of nodes you can pack in a tree with a given height and fan-out T_max(h,f). This way, every other tree T(h,f) is guaranteed to have as much, or less nodes than T_max(h,f). Therefore, if you find such T_max(h,f), that
total_nodes( T_max(h,f) ) > n > total_nodes( T_max(h-1,f))
h will be guaranteed to be the minimum height of a tree with n nodes and f fan-out.
In order to find such a tree, you need to maximize the number of nodes in every layer of a tree. In other words, every node of such a tree needs to have fan-out of f, no less. Therefore you start inserting nodes in a tree, one level at a time. After a layer is full, you start adding another layer. After n nodes are inserted in such a tree, you stop and check the height of the tree. This will be the minimal height you are looking for.
Or, you can do a calculation instead:
nodes_in_level(1) = 1
nodes_in_level(2) = f
nodes_in_level(3) = f * f
...
nodes_in_level(x) = f ^ (x - 1)
This is a standard geometric progression. The maximum nodes of a given tree with height x and fan-out f is therefore the sum of such geometric progression and it shouldn't be too much trouble to figure out the smallest x, for which the number of nodes is greater than n.
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
How do I find the density of a given binary tree? I came across this interview question and not sure as to what they mean by density! Any help would be appreciated.
A dense binary tree is close to perfect (it has close to 2^(h + 1) - 1 nodes). A sparse tree is closer to a linked list (it has close to h nodes). h is the height of the tree where a single root node has height 0.
A simple measure of density could be:
(n - h)/(2^(h + 1) - h - 1)
I just made that formula up, so I don't know if it would suit your needs for an interview answer, but it'll give you 0 for a degenerate tree and 1 for a perfect tree. It will give you numbers close to 1 for dense trees, and numbers close to 0 for sparse ones.
Wikipedia has a lot of information on binary trees.
In a binary trees, the number of nodes at each level falls within a range of values.
At level 0, there is 1 node, the root; at level 1 there can be 1 or 2 nodes.
At any level k, the number of nodes is in the range from 1 to 2k.
The number of nodes per level contributes to the density of the tree. Intuitively, density is a measure of the size of a tree (number of nodes) relative to the height of the tree.