No.
To merge two binary search trees into a single binary search tree, you can perform an in-order traversal on each tree to extract their elements, combine the elements into a single sorted list, and then construct a new binary search tree from the sorted list. This process ensures that the resulting tree maintains the binary search tree property.
There is no such thing. There are binary trees and linked lists.
Linked list was introduced to reduce the space wastage done by array & also to make easier the insertion and deletion of elements from a list. A binary tree contains nodes of elements where insertion,deletion & searching is frequently done. So to make these operations easier linked list is used.
In a linked list, if you want to prevent elements with the same key from being inserted twice, then you need to search the list prior to insert for that key. Iterate through the list, comparing elements with the element being inserted. If you encounter end-of-list, then insert the new element, otherwise throw whatever exception or indicate whatever error desired. Note that this is a full linear search. Statistically, if an element is to be found, it will be found at the halfway point, assuming uniformly random distribution of data. In the worst case, if the element is not found, it will always take a full search to prove that. This is not a very efficient use of a linked-list. It would be better to use some kind of ordered list, perhaps a dynamic array with binary search, or a balanced binary tree, which has similar search performance but one with the most cost to design and implement. You could keep the linked list in order, by inserting each element before the element that has higher key value. This would reduce search time to half, but that is still proportional to list size, not log 2 of list size like binary search is. Sorry, every answer has its tradeoffs.
Binary search requires that the list be in search key order.
* 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)
You compare items in a linked list by searching for them. Iterate through the list, comparing elements with the search key. If you encounter end-of-list, then the key is not found, otherwise you have found the element desired. Note that this is a half linear search. Statistically, if an element is to be found, it will be found at the halfway point, assuming uniformly random distribution of data. In the worst case, if the element is not found, it will always take a full search to prove that. You could keep the linked list in order, by inserting each element before the element that has higher key value. This would reduce search time to half, because searching would stop with the element with higher key value. Searching is not a very efficient use of a linked-list. It would be better to use some kind of ordered list, perhaps a dynamic array with binary search, or a balanced binary tree, which has similar search performance but one with the most cost to design and implement. Linked-lists are better for keeping elements in the order they were encountered or inserted, such as processing tokens in a compiler. Sorry, but every solution has its tradeoffs.
Overcoming the "drawbacks" of a linked list requires knowing what drawback is at stack. If you need to iterate backwards as well as forwards, then you could create a doubly linked list. If you need to search for elements quickly, then you could implement a binary tree. If you have a static size, then you could implement an array. It's all a matter of tradeoff, and of what your particular issue is... Its badsector... According to my self disadvantage of link list that searching in link list is sequential if you compare it with arrays its very slow. Because in link list we have to search every node for that. if any one uses binary tree that is in some cases more faster than arrays.
It's called "Linear Search". If the list is sorted, then it is possible to perform more advanced searches like binary search. If the list isn't sorted, then you can either sort the list first and then binary search or simply use a linear search. Linear search is typically a brute force solution when the data isn't "planned" or if the data is stored in a linked list where random access of the values in the list is slow.
The only drawback I know of is that binary search requires that the list already be sorted. So if you have a really large unsorted list than binary search would not be the best option.
the major limitation of binary search is that there is a need of sorted array to perform binary search operation. if array is not sorted the output is either not correct or may be after a long number of steps and according to data structure the output should come in minimum number of steps.
A tree doesn't do anything so it has no speed...