O(h)
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.
The complexity of the binary search algorithm is log(n)...If you have n items to search, you iteratively pick the middle item and compare it to the search term. Based on that comparision, you then halve the search space and try again. The number of times that you can halve the search space is the same as log2n. This is why we say that binary search is complexity log(n).We drop the base 2, on the assumption that all methods will have a similar base, and we are really just comparing on the same basis, i.e. apples against apples, so to speak.
If the array is unsorted, the complexity is O(n) for the worst case. Otherwise O(log n) using binary search.
Each level of height adds another layer that you must progress through so it is slower.
Average case complexity for Binary search O(log N). (Big O log n)Habibur Rahman (https://www.facebook.com/mmhabib89)BUBT University Bangladeshhttp://www.bubt.edu.bd/
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).
Deezzzz Nutzzzz
In the worst case a binary search tree is linear and has a height equal to the number of nodes. so h=O(h).
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.
The complexity of the binary search algorithm is log(n)...If you have n items to search, you iteratively pick the middle item and compare it to the search term. Based on that comparision, you then halve the search space and try again. The number of times that you can halve the search space is the same as log2n. This is why we say that binary search is complexity log(n).We drop the base 2, on the assumption that all methods will have a similar base, and we are really just comparing on the same basis, i.e. apples against apples, so to speak.
both seach has different algorithem but the complexity will be same...
Check this out! http://stackoverflow.com/questions/575772/the-best-way-to-calculate-the-height-in-a-binary-search-tree-balancing-an-avl
If the array is unsorted, the complexity is O(n) for the worst case. Otherwise O(log n) using binary search.
Binary search is used for large arrays because it is the fastest search, on the order of O-Log2-N complexity, which means that the maximum number of compare operations to find a specific item is Log2N, where N is the number of elements.
Each level of height adds another layer that you must progress through so it is slower.
Average case complexity for Binary search O(log N). (Big O log n)Habibur Rahman (https://www.facebook.com/mmhabib89)BUBT University Bangladeshhttp://www.bubt.edu.bd/
For the height `h' of a binary tree, for which no further attributes are given than the number `n' of nodes, holds:ceil( ld n)