the successor for a binary search tree is O(n) in the worst case and O( log n) i
ID: 3644248 • Letter: T
Question
the successor for a binary search tree is O(n) in the worst case and O(log n) in the typical case. Suppose the successor operation is used to implement an iterator that will return the elements in ascending order, starting from the minimum element of the tree.1 Explain why the amortized cost of the successor operation, when called n times to traverse the elements in order, is O(1). (Hints: compare the execution of the successor operation with the execution of a recursive inorder traversal.) let assume that the successor operation is a standard implementation.
Explanation / Answer
A splay tree is a self-adjusting binary search tree with the additional property that recently accessed elements are quick to access again. It performs basic operations such as insertion, look-up and removal in O(log n) amortized time. For many sequences of nonrandom operations, splay trees perform better than other search trees, even when the specific pattern of the sequence is unknown. Element provides a method traverse that traverses the tree and calls user-specified callbacks for each node, in pre- or post-order. However, use of the method is quite superfluous: if you want to recursively visit every node in the tree, it's almost always simpler to write a subroutine does just that, than it is to bundle up the pre- and/or post-order code in callbacks for the traverse method. { my $counter = 'x0000'; $start_node->traverse( [ # Callbacks; # pre-order callback: sub { my $x = $_[0]; $x->attr('id', $counter++) unless defined $x->attr('id'); return HTML::Element::OK; # keep traversing }, # post-order callback: undef ], 1, # don't call the callbacks for text nodes ); } The traverse() method is a general object-method for traversing a tree or subtree and calling user-specified callbacks.
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