The complexity of binary search tree :
Search , Insertion and Deletion is O(h) . and the Height can be of O(n) ( if the tree is a skew tree).
For Balanced Binary Trees , the Order is O(log n).
O(log n)At each step of insertion you are either going to the left child or the right child. In a balanced tree, this will effectively cut the number of possible comparisons in half each time.
A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible.
yes, why not,
Advantages:BST is fast in insertion and deletion etc when balanced.Very efficient and its code is easier than link lists.Disadvantages:Shape of the tree depends upon order of insertion and it can be degenerated.Searching takes long time.
The best case for a binary search is finding the target item on the first look into the data structure, so O(1). The worst case for a binary search is searching for an item which is not in the data. In this case, each time the algorithm did not find the target, it would eliminate half the list to search through, so O(log n).
O(h)
Deezzzz Nutzzzz
no they are not same
Binary search is a log n type of search, because the number of operations required to find an element is proportional to the log base 2 of the number of elements. This is because binary search is a successive halving operation, where each step cuts the number of choices in half. This is a log base 2 sequence.
By using Depth First Search or Breadth First search Tree traversal algorithm we can print data in Binary search tree.
self depend friend"s............
* search array => O(1) linked list=> O(n) binary tree=> O(log n) hash=>O(1) * search array => O(1) linked list=> O(n) binary tree=> O(log n) hash=>O(1)
A binary search tree is already ordered. An in order traversal will give you a sorted list of nodes.
Binary trees are commonly used to implement binary search tree and binary heaps.
a tree which has atmost two nodes is called binary tree binary search tree is a binary tree which satisfies the following 1.every node in tree must be distinct 2.values in right subtree > value at root 3.values in left subtree < value at root 4.left,right subtrees must be binary search trees
O(log n)At each step of insertion you are either going to the left child or the right child. In a balanced tree, this will effectively cut the number of possible comparisons in half each time.
Well, you might if you want to.