14. Binary Search Trees

Written by Kelvin Lau

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:

That’s why array.contains(_:) is an O(n) operation.

This is not the case for binary search trees:

Every time the search algorithm visits a node in the BST, it can safely make these two assumptions:

• If the search value is less than the current value, it must be in the left subtree.
• If the search value value is greater than the current value, it must be in the right subtree.

By leveraging the rules of the BST, you can avoid unnecessary checks and cut the search space in half every time you make a decision. That’s why element lookup in a BST is an O(log n) operation.

Insertion

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

Inserting values into an array is like butting into an existing line: Everyone in the line behind your chosen spot needs to make space for you by shuffling back.

In the above example, zero is inserted in front of the array, causing all other elements to shift backwards by one position. Inserting into an array has a time complexity of O(n).

Insertion into a binary search tree is much more comforting:

By leveraging the rules for the BST, you only needed to make three traversals to find the location for the insertion, and you didn’t have to shuffle all the elements around! Inserting elements in a BST is again an O(log n) operation.

Removal

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

This behavior also plays nicely with the lineup analogy. If you leave the middle of the line, everyone behind you needs to shuffle forward to take up the empty space.

Here’s what removing a value from a BST looks like:

Nice and easy! There are complications to manage when the node you’re removing has children, but you’ll look into that later. Even with those complications, removing an element from a BST is still an O(log n) operation.

Binary search trees drastically reduce the number of steps for add, remove and lookup operations. Now that you have a grasp of the benefits of using a binary search tree, you can move on to the actual implementation.

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)
  }
}

By definition, binary search trees can only hold values that are Comparable.

Note: You could relax the requirement for Comparable by using closures for comparison. We will stick with the Comparable because it is simpler and allows us to focus on the core concepts.

Next, you’ll look at the insert method.

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.

Add the following to BinarySearchTree.swift:

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
  }
}

The first insert method is exposed to users, while the second one will be used as a private helper method:

1. This is a recursive method, so it requires a base case for terminating recursion. If the current node is nil, you’ve found the insertion point and you return the new BinaryNode.

2. Because Element types are comparable, you can perform a comparison. This if statement controls which way the next insert call should traverse. If the new value is less than the current value, you call insert on the left child. If the new value is greater than or equal to the current value, you’ll call insert on the right child.

3. Return the current node. This makes assignments of the form node = insert(from: node, value: value) possible as insert will either create node (if it was nil) or return node (it it was not nil).

Head back to the playground page and add the following at the bottom:

example(of: "building a BST") {
  var bst = BinarySearchTree<Int>()
  for i in 0..<5 {
    bst.insert(i)
  }
  print(bst)
}

You should see the following output:

---Example of: building a BST---
    ┌──4
  ┌──3
  │ └──nil
 ┌──2
 │ └──nil
┌──1
│ └──nil
0
└──nil

That tree looks a bit unbalanced, but it does follow the rules. However, this tree layout has undesirable consequences. When working with trees, you always want to achieve a balanced format:

An unbalanced tree affects performance. If you insert 5 into the unbalanced tree you’ve created, it becomes an O(n) operation:

You can create structures known as self-balancing trees that use clever techniques to maintain a balanced structure, but we’ll save those details for Chapter 16, “AVL Trees”. For now, you’ll simply build a sample tree with a bit of care to keep it from becoming unbalanced.

Add the following computed variable at the top of the playground page:

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
}

Replace your example function with the following:

example(of: "building a BST") {
  print(exampleTree)
}

You should see the following in the console:

---Example of: building a BST---
 ┌──5
┌──4
│ └──nil
3
│ ┌──2
└──1
 └──0

Much nicer!

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.

Add the following to the bottom of BinarySearchTree.swift:

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
  }
}

Next, head back to the playground page to test this out:

example(of: "finding a node") {
  if exampleTree.contains(5) {
    print("Found 5!")
  } else {
    print("Couldn’t find 5")
  }
}

You should see the following in the console:

---Example of: finding a node---
Found 5!

In-order traversal has a time complexity of O(n), thus this implementation of contains has the same time complexity as an exhaustive search through an unsorted array. However, you can do better.

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
}

1. Start by setting current to the root node.
2. While current is not nil, check the current node’s value.
3. If the value is equal to what you’re trying to find, return true.
4. Otherwise, decide whether you’re going to check the left or the right child.

This implementation of contains is an O(log n) operation in balanced binary search tree.

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.

For non-leaf nodes, however, there are extra steps you must take.

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:

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:

Simply deleting the node presents a dilemma:

You have two child nodes (12 and 37) to reconnect, but the parent node only has space for one child. To solve this problem, you’ll implement a clever workaround by performing a swap.

When removing a node with two children, replace the node you removed with smallest node in its right subtree. Based on the rules of the BST, this is the leftmost node of the right subtree:

It’s important to note that this produces a valid binary search tree. Because the new node was the smallest node in the right subtree, all nodes in the right subtree will still be greater than or equal to the new node. And because the new node came from the right subtree, all nodes in the left subtree will be less than the new node.

After performing the swap, you can simply remove the value you copied, which is just a leaf node.

This will take care of removing nodes with two children.

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
  }
}

This should look familiar to you. You’re using the same recursive setup with a private helper method as you did for insert. You’ve also added a recursive min property to BinaryNode to find the minimum node in a subtree. The different removal cases are handled in the if value == node.value clause:

// 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)

1. In the case in which the node is a leaf node, you simply return nil, thereby removing the current node.
2. If the node has no left child, you return node.rightChild to reconnect the right subtree.
3. If the node has no right child, you return node.leftChild to reconnect the left subtree.
4. This is the case in which the node to be removed has both a left and right child. You replace the node’s value with the smallest value from the right subtree. You then call remove on the right child to remove this swapped value.

Head back to the playground page and test remove by writing the following:

example(of: "removing a node") {
  var tree = exampleTree
  print("Tree before removal:")
  print(tree)
  tree.remove(3)
  print("Tree after removing root:")
  print(tree)
}

You should see the following output in the console:

---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.