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)
O(h)
Binary search requires that the list be in search key order.
A binary search is much faster.
(i) Binary search can interact poorly with the memory hierarchy (i.e. caching), because of its random-access nature. For in-memory searching, if the interval to be searching is small, a linear search may have superior performance simply because it exhibits better locality of reference. (ii) Binary search algorithm employs recursive approach and this approach requires more stack space. (iii) Programming binary search algorithm is very difficult and error prone (Kruse, 1999).
In the worst case a binary search tree is linear and has a height equal to the number of nodes. so h=O(h).
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).
Check this out! http://stackoverflow.com/questions/575772/the-best-way-to-calculate-the-height-in-a-binary-search-tree-balancing-an-avl
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)
O(h)
Binary Search trees offer improved average case performance for searching. However, if they are unbalanced their search performance degrades to that of a linked list. AVL trees guarantee that the difference in height of any two subtrees rooted at the same node will be at most one. This guarantees an asymptotic running time of O(log(n)) as opposed to O(n) in the case of a standard bst.
Binary search requires that the list be in search key order.
Binary trees are commonly used to implement binary search tree and binary heaps.
A binary search is much faster.
The only items suitable for a binary search are those which are in a sorted order.
It is 10111111 in binary. Try a search for '191 to binary'.
no they are not same