CS106X – Programming Abstractions in C++ Cynthia Bailey Lee CS2 in C++ Peer Instruction Materials by Cynthia Bailey Lee is licensed under a Creative Commons Attribution-NonCommercial- ShareAlike 4.0 International License. Permissions beyond the scope of this license may be available at Bailey LeeCreative Commons Attribution-NonCommercial- ShareAlike 4.0 International Licensehttp://peerinstruction4cs.org

Today’s Topics: Map Interface, continued 1. Binary Search Tree math  (continued from Friday)  How to maintain balance 2. Introduction to Hashing 3. C++ template details (if we have time) 2

BST math

Bad BSTs  One way to create a bad BST is to insert the elements in order: 34, 22, 18, 9, 3  That’s not the only way…  How many distinctly structured BSTs are there that exhibit the worst case height (height equals number of nodes) for a tree with 5 nodes? A. 2-3 B. 4-5 C. 6-7 D. 8-9 E. More than 9

BALANCE!!!  The #1 issue to remember with BSTs is that they are great when balanced (O(log n) operations), and horrible when unbalanced (O(n) operations)  Balance depends on order of insert of elements  Not the implementor’s control  Over the years, implementors have devised ways of making sure BSTs stay balanced no matter what order the elements are inserted

BST Balance Strategies

Red-Black trees One of the most famous (and most tricky) strategies for keeping a BST balanced

Red-Black trees  In addition to the requirements of binary search trees, red–black trees must meet these:  A node is either red or black.  The root is black.  All leaves (null children) are black.  Both children of every red node are black.  Every simple path from a given node to any of its descendant leaves contains the same number of black nodes.

Red-Black trees  In addition to the requirements of binary search trees, red–black trees must meet these:  A node is either red or black.  The root is black.  All leaves (null children) are black.  Both children of every red node are black.  Every simple path from a given node to any of its descendant leaves contains the same number of black nodes. According to these, what is the greatest possible difference between the longest and shortest root-to-leaf paths in the tree? A.All root-to-leaf paths are equal length B.25% longer C.50% longer D.100% longer E.Other/none/more than one

Red-Black trees  Every simple path from a given node to any of its descendant leaves contains the same number of black nodes.  This is what guarantees “close” to balance This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license. CommonsAttribution-Share Alike 3.0 Unported

Insert procedure must maintain the invariants (this gets tricky)  Video: This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license. CommonsAttribution-Share Alike 3.0 Unported

Other BST balance strategies  Red-Black tree  AVL tree  Treap (BST + heap in one tree! What could be cooler than that, amirite? ♥ ♥ ♥ ) Other fun types of BST :  Splay tree  Rather than only worrying about balance, Splay Tree dynamically readjusts based on how often users search for an item. Commonly-searched items move to the top, saving time  B-Tree  Like BST, but a node can have many children, not just 2

Hashing Implementing the Map interface with Hashing/Hash Tables

Imagine you want to look up your neighbors’ names, based on their house number (all on your same street)  House numbers: through  (roughly 1000 houses—there are varying gaps in house numbers between houses)  Names: one last name per house

Options  BST (balanced):  Add/remove: O(logn)  Find: O(logn)  Linked list :  Add/remove: O(n)  Find: O(n)  Array :  Add/remove:  Find: Indexvalue 0“” 1 …… 10565“Kyung Lee” 10566“” 10567“Isaiah White” …… 90598“” 90599“” 90600“Solange Clark”

Hash Table is just a modified, more flexible array  Keys don’t have to be integers 0-(size-1)  (Ideally) avoids big gaps like our gap from 0 to in the house numbers and between numbers  Hash function is what makes this possible!

Closed Addressing Array index Value  Where does key=“Annie” value=10 go if hash (“Annie”) = 3?  Where does key=“Solange” value=12 go if hash(“Solange”) = 5?

Hash collisions  We may need to worry about hash collisions  Hash collision:  Two keys a, b, a≠b, have the same hash code index (i.e. hash(a) = hash(b))

Closed Addressing Array index Value  Where does key=“Annie” value=55 go if hashkey(“Annie”) = 3? A. 55 overwrites 10 at 3 B. A link list node is added at 3 C. Other/none/ more than one

Hash key collisions  We can NOT overwrite the value the way we would if it really were the same key  Need a way of storing multiple values in a given “place” in the hash table