If I were to have a modified version of a binary tree such that the maximal difference between the right and left child nodes of every node in the tree is ≤2, my intuition tells me that similarly to a regular binary tree, this would have a height of O(log n), but how can I prove this using induction?
Related
What is the time complexity of doing O(h) algorithm when h is the height of the node in BST n times (the number of elements in the tree) , I believe it's O(n) and not O(n*h) but I have no clue how to prove it.
The specific algorithm that works in O(h) is finding the In-order predecessor of an element in BST.
The cost of computing inorder successors n times in any BST is O(n). To see this, count how many times you touch each edge in the tree. You’ll pass down the edge once when you first explore a subtree, and once more after you leave it. Overall, this means you touch each edge at most twice, so the total work done is O(n).
Note that generally speaking you can upper-bound the cost of performing n O(h)-time on a BST that has height h at O(hn), and that will never underestimate things. However, if you know more specifically about the algorithm you’re using, as in this case, you can get a tighter bound.
O(n²).
A binary search tree is not balanced, which means that the height of a node can be equal to the number of nodes of the tree, hence O(n²).
Can anyone mathematically prove me why the following BST definition is always true by using the fact that the height of a BST is always logN where N is the number of nodes.
"a binary tree in which the left and right subtrees of every node differ in height by no more than 1."
I have an one lecture slides says following:
To find middle element in AVL tree, I traverse elements in order until It reaches the moddile element. It takes O(N).
If I know correctly, in tree structure, finding element takes base 2 O(logn) since AVL is binary tree that always divided into 2 childs.
But why it says O(N)?
I am just trying to elaborate 'A. Mashreghi' comment.
Since, the tree under consideration is AVL tree - the guaranteed finding of element in O(log n) holds as log as you have the element(key) to find.
The problem is - you are trying to identify a middle element in the given data structure. As it is AVL tree (self balanced BST) in-order travel gives you elements in ascending order. You want to use this property to find the middle element.
Algorithm goes like - have a counter increment for every node traversed in-order and return # n/2th position. This sums to O(n/2) and hence the overall complexity O(n).
Being divided into 2 children does not guarantee perfect symmetry. For instance, consider the most unbalanced of all balanced binary trees: each right child has a depth one more than its corresponding left child.
In such a tree, the middle element will be somewhere down in the right branch's left branch's ...
You need to determine how many nodes N you have, then locate the N/2th largest node. This is not O(log N) process.
Would a balanced binary search tree help you complete the following task in a faster big-oh time than a balanced binary tree?
Creating a list of all elements in the tree that are smaller than some value v.
In my opinion no because what if all the values in the BST are smaller than v. Then you would have to visit each node and that would be O(n) which is not any better than a binary tree.
Am I correct?
In my opinion no because what if all the values in the BST are smaller
than v. Then you would have to visit each node and that would be O(n)
which is not any better than a binary tree.
Am I correct?
You are. Notice however that for all practical purposes is better to use BST, because with "plain" binary tree you always have to visit all nodes in order to find those smaller than v, while in BST with in-order traversal you examine only those smaller than v.
What is the name of the binary tree (or the family of the binary
trees), that is balanced, and has the minimum number of nodes
possible for its height?
balanced binary tree
(data structure)
Definition: A binary tree where no leaf is more than a certain amount farther from the root than any other. After inserting or deleting a node, the tree may rebalanced with "rotations."
Generalization (I am a kind of ...)
binary tree.
Specialization (... is a kind of me.)
AVL tree, red-black tree, B-tree, balanced binary search tree.
Aggregate child (... is a part of or used in me.)
left rotation, right rotation.
See also BB(α) tree, height-balanced tree.
-- http://www.itl.nist.gov/div897/sqg/dads/HTML/balancedbitr.html
It is called Fibonacci tree
AVL is a balanced tree with log(n) height (this is the lowest height possible for binary tree).
Another implementation of a similar data structure is Red Black Tree.
Both trees implement all operations in O(log(n)).
AVL Tree is something that you have been looking for.
From Wikipedia:
In computer science, an AVL tree is a self-balancing binary search tree, and it is the first such data structure to be invented. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; therefore, it is also said to be height-balanced. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations.