17. AVL Tree Challenges

Written by Kelvin Lau

Here are three challenges that revolve around AVL trees. Solve these to make sure you have the concepts down.

Challenge 1: Number of leaves

How many leaf nodes are there in a perfectly balanced tree of height 3? What about a perfectly balanced tree of height h?

Challenge 2: Number of nodes

How many nodes are there in a perfectly balanced tree of height 3? What about a perfectly balanced tree of height h?

Challenge 3: A tree traversal protocol

Since there are many variants of binary trees, it makes sense to group shared functionality in a protocol. The traversal methods are a good candidate for this.

Create a TraversableBinaryNode protocol that provides a default implementation of the traversal methods so that conforming types get these methods for free. Have AVLNode conform to this.

Note: You’ll need to make AVLNode a final class because the protocol will have Self requirements.

Solutions

Solution to Challenge 1

A perfectly balanced tree is a tree where all the leaves are in the same level, and that level is completely filled:

Recall that a tree with just a root node has a height of zero. Thus, the tree in the example above has a height of two. You can extrapolate that a tree with a height of three would have eight leaf nodes.

Since each node has two children, the number of leaf nodes doubles as the height increases. You can calculate the number of leaf nodes with a simple equation:

func leafNodes(inTreeOfHeight height: Int) -> Int {
  Int(pow(2.0, Double(height)))
}

Solution to Challenge 2

Since the tree is perfectly balanced, the number of nodes in a perfectly balanced tree of height 3 can be expressed by the following:

func nodes(inTreeOfHeight height: Int) -> Int {
  var totalHeight = 0
  for currentHeight in 0...height {
    totalHeight += Int(pow(2.0, Double(currentHeight)))
  }
  return totalHeight
}

Although this certainly gives you the correct answer, there is a faster way. If you examine the results of a sequence of height inputs, you’ll realize that the total number of nodes is one less than the number of leaf nodes of the next level.

Thus, a faster version of this is the following:

func nodes(inTreeOfHeight height: Int) -> Int {
  Int(pow(2.0, Double(height + 1))) - 1
}

Solution to Challenge 3

First, create the following protocol:

protocol TraversableBinaryNode {

  associatedtype Element
  var value: Element { get }
  var leftChild: Self? { get }
  var rightChild: Self? { get }
  func traverseInOrder(visit: (Element) -> Void)
  func traversePreOrder(visit: (Element) -> Void)
  func traversePostOrder(visit: (Element) -> Void)
}

Then add a protocol extension to provide implementations for the traversal methods:

extension TraversableBinaryNode {

  func traverseInOrder(visit: (Element) -> Void) {
    leftChild?.traverseInOrder(visit: visit)
    visit(value)
    rightChild?.traverseInOrder(visit: visit)
  }

  func traversePreOrder(visit: (Element) -> Void) {
    visit(value)
    leftChild?.traversePreOrder(visit: visit)
    rightChild?.traversePreOrder(visit: visit)
  }

  func traversePostOrder(visit: (Element) -> Void) {
    leftChild?.traversePostOrder(visit: visit)
    rightChild?.traversePostOrder(visit: visit)
    visit(value)
  }
}

Next, head into AVLNode.swift update the type declaration to include the final keyword:

public final class AVLNode<Element>

Finally, add the following at the bottom of the playground:

extension AVLNode: TraversableBinaryNode {}

example(of: "using TraversableBinaryNode") {
  var tree = AVLTree<Int>()
  for i in 0..<15 {
    tree.insert(i)
  }
  tree.root?.traverseInOrder { print($0) }
}

Verify that you’re getting the following results in the console:

---Example of: using TraversableBinaryNode---
0
1
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3
4
5
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7
8
9
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14