Home iOS & Swift Books Data Structures & Algorithms in Swift

14
Binary Search Trees Written by Kelvin Lau

Heads up... You're reading this book for free, with parts of this chapter shown beyond this point as scrambled text.

You can unlock the rest of this book, and our entire catalogue of books and videos, with a raywenderlich.com Professional subscription.

A binary search tree, or BST, is a data structure that facilitates fast lookup, insert and removal operations. Consider the following decision tree where picking a side forfeits all the possibilities of the other side, cutting the problem in half.

Once you make a decision and choose a branch there is no looking back. You keep going until you make a final decision at a leaf node. Binary trees let you do the same thing. Specifically, a binary search tree imposes two rules on the binary tree you saw in the previous chapter:

  • The value of a left child must be less than the value of its parent.
  • Consequently, the value of a right child must be greater than or equal to the value of its parent.

Binary search trees use this property to save you from performing unnecessary checking. As a result, lookup, insert and removal have an average time complexity of O(log n), which is considerably faster than linear data structures such as arrays and linked lists.

In this chapter, you’ll learn about the benefits of the BST relative to an array and as usual, implement the data structure from scratch.

Case study: array vs. BST

To illustrate the power of using a BST, you’ll look at some common operations and compare the performance of arrays against the binary search tree.

Consider the following two collections:

Lookup

There’s only one way to do element lookups for an unsorted array. You need to check every element in the array from the start:

Searching for 105
Wiidvxalw zov 906

Searching for 105
Baupdyimt fez 128

Insertion

The performance benefits for the insertion operation follow a similar story. Assume you want to insert 0 into a collection:

Inserting 0 in sorted order
Opsunlujz 7 ed hansuy untut

Removal

Similar to insertion, removing an element in an array also triggers a shuffling of elements:

Removing 25 from the array
Gorekikg 89 lzul fcu azmac

Implementation

Open up the starter project for this chapter. In it, you’ll find the BinaryNode type that you created in the previous chapter. Create a new file named BinarySearchTree.swift and add the following inside the file:

public struct BinarySearchTree<Element: Comparable> {

  public private(set) var root: BinaryNode<Element>?

  public init() {}
}

extension BinarySearchTree: CustomStringConvertible {

  public var description: String {
    guard let root = root else { return "empty tree" }
    return String(describing: root)
  }
}

Inserting elements

In accordance with the rules of the BST, nodes of the left child must contain values less than the current node. Nodes of the right child must contain values greater than or equal to the current node. You’ll implement the insert method while respecting these rules.

extension BinarySearchTree {

  public mutating func insert(_ value: Element) {
    root = insert(from: root, value: value)
  }
  
  private func insert(from node: BinaryNode<Element>?, value: Element)
      -> BinaryNode<Element> {
    // 1
    guard let node = node else {
      return BinaryNode(value: value)
    }
    // 2
    if value < node.value {
      node.leftChild = insert(from: node.leftChild, value: value)
    } else {
      node.rightChild = insert(from: node.rightChild, value: value)
    }
    // 3
    return node
  }
}
example(of: "building a BST") {
  var bst = BinarySearchTree<Int>()
  for i in 0..<5 {
    bst.insert(i)
  }
  print(bst)
}
---Example of: building a BST---
    ┌──4
  ┌──3
  │ └──nil
 ┌──2
 │ └──nil
┌──1
│ └──nil
0
└──nil

var exampleTree: BinarySearchTree<Int> {
  var bst = BinarySearchTree<Int>()
  bst.insert(3)
  bst.insert(1)
  bst.insert(4)
  bst.insert(0)
  bst.insert(2)
  bst.insert(5)
  return bst
}
example(of: "building a BST") {
  print(exampleTree)
}
---Example of: building a BST---
 ┌──5
┌──4
│ └──nil
3
│ ┌──2
└──1
 └──0

Finding elements

Finding an element in a BST requires you to traverse through its nodes. It’s possible to come up with a relatively simple implementation by using the existing traversal mechanisms that you learned about in the previous chapter.

extension BinarySearchTree {

  public func contains(_ value: Element) -> Bool {
    guard let root = root else {
      return false
    }
    var found = false
    root.traverseInOrder {
      if $0 == value {
        found = true
      }
    }
    return found
  }
}
example(of: "finding a node") {
  if exampleTree.contains(5) {
    print("Found 5!")
  } else {
    print("Couldn’t find 5")
  }
}
---Example of: finding a node---
Found 5!

Optimizing contains

You can rely on the rules of the BST to avoid needless comparisons. Back in BinarySearchTree.swift, update the contains method to the following:

public func contains(_ value: Element) -> Bool {
  // 1
  var current = root
  // 2
  while let node = current {
    // 3
    if node.value == value {
      return true
    }
    // 4
    if value < node.value {
      current = node.leftChild
    } else {
      current = node.rightChild
    }
  }
  return false
}

Removing elements

Removing elements is a little more tricky, as there are a few different scenarios you need to handle.

Case 1: Leaf node

Removing a leaf node is straightforward; simply detach the leaf node.

removing 2
vukadexz 5

Case 2: Nodes with one child

When removing nodes with one child, you’ll need to reconnect that one child with the rest of the tree:

removing 4, which has 1 child
sokagicg 9, bxojr lem 3 dyogk

Case 3: Nodes with two children

Nodes with two children are a bit more complicated, so a more complex example tree will serve better to illustrate how to handle this situation. Assume that you have the following tree and that you want to remove the value 25:

Implementation

Open up BinarySearchTree.swift to implement remove. Add the following code at the bottom of the file:

private extension BinaryNode {

  var min: BinaryNode {
    leftChild?.min ?? self
  }
}

extension BinarySearchTree {
  
  public mutating func remove(_ value: Element) {
    root = remove(node: root, value: value)
  }
  
  private func remove(node: BinaryNode<Element>?, value: Element)
    -> BinaryNode<Element>? {
    guard let node = node else {
      return nil
    }
    if value == node.value {
      // more to come
    } else if value < node.value {
      node.leftChild = remove(node: node.leftChild, value: value)
    } else {
      node.rightChild = remove(node: node.rightChild, value: value)
    }
    return node
  }
}
// 1
if node.leftChild == nil && node.rightChild == nil {
  return nil
}
// 2
if node.leftChild == nil {
  return node.rightChild
}
// 3
if node.rightChild == nil {
  return node.leftChild
}
// 4
node.value = node.rightChild!.min.value
node.rightChild = remove(node: node.rightChild, value: node.value)
example(of: "removing a node") {
  var tree = exampleTree
  print("Tree before removal:")
  print(tree)
  tree.remove(3)
  print("Tree after removing root:")
  print(tree)
}
---Example of: removing a node---
Tree before removal:
 ┌──5
┌──4
│ └──nil
3
│ ┌──2
└──1
 └──0

Tree after removing root:
┌──5
4
│ ┌──2
└──1
 └──0

Key points

  • The binary search tree is a powerful data structure for holding sorted data.
  • Elements of the binary search tree must be comparable. This can be achieved using a generic constraint or by supplying closures to compare with.
  • The time complexity for insert, remove and contains methods in a BST is O(log n).
  • Performance will degrade to O(n) as the tree becomes unbalanced. This is undesirable, so you’ll learn about a self-balancing binary search tree called the AVL tree in Chapter 16.

Have a technical question? Want to report a bug? You can ask questions and report bugs to the book authors in our official book forum here.

Have feedback to share about the online reading experience? If you have feedback about the UI, UX, highlighting, or other features of our online readers, you can send them to the design team with the form below:

© 2021 Razeware LLC

You're reading for free, with parts of this chapter shown as scrambled text. Unlock this book, and our entire catalogue of books and videos, with a raywenderlich.com Professional subscription.

Unlock Now

To highlight or take notes, you’ll need to own this book in a subscription or purchased by itself.