CSCI H343 Data Structures Fall 2023

More Abstract Data Types

Stack (LIFO)

analogy: stack of pancakes

interface Stack<E> {
	void push(E d);
	E pop();
	E peek();
	boolean empty();
}

Example uses:

• reverse an array
• check for matching parentheses (with multiple kinds: square, round, curly)
• parsing (e.g. HTML)
• depth-search search
• procedure call stack

Implementation:

• array (push is add to the end, pop removes from end)
• singly-linked list (push on front, pop from front)

Queue (FIFO)

Analogy: checkout line at a grocery store

interface Queue<E> {
	void enqueue(E e); // aka. push
	E dequeue(); // aka. pop
	E front();
	boolean empty();
}

Example uses:

• breadth-first search,
• requests to a shared resource (e.g., printer),
• interupt handling inside an OS kernel,
• buffering to handle asynchronous communiation, such as interprocess IO, disk access, etc.

Implementation:

• array
• doubly-linked list

Set

Like a set in mathematics. A collection of elements where the ordering of the elements is not important, only membership matters. Set ignores duplicates.

interface Set {
   void insert(int e);
   void remove();
   boolean member(int e);
   boolean empty();
   Set union(Set other);
   Set intersect(Set other);
   Set difference(Set other);
}

Implementations of Set are the topic of several future lectures.

Binary Search Trees

Motivation

Taking stock with respect to Flood-it! and the Set ADT

	array	linked list	sorted array
membership test	O(n)	O(n)	O(log(n))
insertion	O(1)	O(1)	O(n)

• How do we keep the array sorted when inserting more elements?

  Use binary search for insertion.

  What’s the time complexity?

  O(n) because we still have to shift elements down.

• We’d like something that can do both membership test and insertion in O(lg n).

• Can we create a hybrid of the sorted array and the linked list?

  Yes: binary search trees!

• This lecture we discuss binary trees as a lead-up to discussing binary search trees in the next lecture.

Binary Node and Tree Classes

class BinaryNode<T> {
	T data;
	BinaryNode<T> left;
	BinaryNode<T> right;

	BinaryNode(T d, BinaryNode<T> l, BinaryNode<T> r) {
			data = d; left = l; right = r;
	}
}

public class BinaryTree<T> {
	BinaryNode<T> root;

	public BinaryTree() { root = null; }
	// ...
}

We can traverse a binary tree in several different ways:

• preorder: me, left, right
• inorder: left, me, right
• postorder: left, right, me

Example:

             10
            / \
           /   \
          5     14
         / \     \
        2   7     19

        pre:  10,5,2,7,14,19
        in:   2,5,7,10,14,19
        post: 2,7,5,19,14,10

Next and Previous operations with respect to inorder traversal

Running example binary tree:

      8
    /  \
   /    \
  3     10
 / \      \
1   6      14
   / \    /
  4   7  13

Helper functions first and last

Find the first node according to an inorder traversal within a subtree.

Examples:

What’s the first node in the above tree? answer: 1

What’s the first node of the subtree rooted at 6? answer: 4

public BinaryNode<T> first();

Algorithm: Go to the left as far as you can.

Find the last node according to an inorder traversal.

public BinaryNode<T> last();

Algorithm: Go to the right as far as you can.

Helper functions nextAncestor and prevAncestor

Given a node, find the ancestor that would come next according to an inorder traversal, or return null if there is none.

You may assume that each node has a parent attribute.

Student in-class exercise:

BinaryNode<T> nextAncestor();

keep going up so long as the node is the right child, when left child, return parent

Solution:

BinaryNode<T> nextAncestor() {
	if (parent != null && this == parent.right) {
		return parent.nextAncestor();
	} else {
		return parent;
	}
}

Given a node, find the ancestor that would come before the node according to an inorder traversal, or return null if there is none.

BinaryNode<T> prevAncestor();

How to find the next node according to an inorder traversal?

Examples:

What is after 3? Answer: 4.

What is after 7? Answer: 8.

If their is a right child, get the first node in the subtree. Otherwise, find the next ancestor.

public BinaryNode<T> next() {
	if (right == null) {
	   return nextAncestor();
	} else {
	   return right.first();
	}
}

What is the time complexity? answer: $O(h)$ where $h$ is the height of the tree.

Finding the previous node is similar.