Practice with Recursion : 50
Trees and Recursion : 49
Trees : 48
Recursion : 47
Lists Review and Performance : 46
Linked Lists : 45
Algorithms and Lists : 44
Lambda Expressions : 43
Anonymous Classes : 42
Practice with Interfaces : 41
Implementing Interfaces : 40
Using Interfaces : 39
Working with Exceptions : 38
Throwing Exceptions : 37
Catching Exceptions : 36
References and Polymorphism : 35
References : 34
Data Modeling 2 : 33
Equality and Object Copying : 32
Polymorphism : 31
Inheritance : 30
Data Modeling 1 : 29
Static : 28
Encapsulation : 27
Constructors : 26
Objects, Continued : 25
Introduction to Objects : 24
Compilation and Type Inference : 23
Practice with Collections : 22
Maps and Sets : 21
Lists and Type Parameters : 20
Imports and Libraries : 19
Multidimensional Arrays : 18
Practice with Strings : 17
null : 16
Algorithms and Strings : 15
Strings : 14
Functions and Algorithms : 13
Practice with Functions : 12
More About Functions : 11
Errors and Debugging : 10
Functions : 9
Practice with Loops and Algorithms : 8
Algorithms I : 7
Loops : 6
Arrays : 5
Compound Conditionals : 4
Conditional Expressions and Statements : 3
Operations on Variables : 2
Variables and Types : 1
Hello, world! : 0
Trees and Recursion
import cs125.trees.BinaryTree;
int countLeftGreaterThanRight(BinaryTree tree) {
return 0;
}
assert countLeftGreaterThanRight(new BinaryTree<Integer>(0, 1, 2)) == 1;
Next we’ll continue practicing with trees and recursion!
And what better way to do that then to do a few problems together?
So let’s get started!
As a warm up, let’s do another counting problem.
Given a binary tree containing Integer
s, let’s count the number of nodes where the value of the left child is greater than the value of the right child.
Before we start, remember the core of our approach to recursion:
- Identify the base case—the simplest problem that you have to be able to immediately solve
- Make the problem smaller at each step
- Combine results appropriately
Interactive Walkthrough
Click on an icon below or the play button above to start!
Next, we’ll look at how to determine if a binary tree contains a certain value.
This problem introduces a new wrinkle to our usual approach to recursion!
Interactive Walkthrough
Click on an icon below or the play button above to start!
Solve: Binary Tree Count Equal to Child (Practice)
Created By: Geoffrey Challen
/ Version: 2020.11.0
Create a public class BinaryTreeCounter
that provides a single class method named countEqualToEitherChild
that
accepts a single BinaryTree
and counts the number of nodes in the tree where the value at that node is equal to
either the value at its right child or the value at its left child. Keep in mind that not every node has a
right or left child, so you'll need to check for null
carefully. (Or use try-catch
!) However, you can assume
that all of the values in the tree are non-null.
For reference, cs125.trees.BinaryTree
has the following public properties:
Next, let’s examine the performance of our recursive algorithms, and determine what O(n) category they belong in.
Interactive Walkthrough
Click on an icon below or the play button above to start!
Solve: BinaryTree Count Equal Children
Created By: Geoffrey Challen
/ Version: 2021.10.0
Create a public class BinaryTreeCounter
that provides a single class method named countEqualChildren
that
accepts a single BinaryTree<?>
and counts the number of nodes in the tree that have two children with equal
values.
Keep in mind that not every node has a right or left child, so you'll need to check for null
carefully.
(Or use try-catch
!)
However, you can assume that all the values in the tree are non-null.
For reference, cs125.trees.BinaryTree
has the following public properties:
More Practice
Need more practice? Head over to the practice page.