# 14 Binary Search Trees Written by Kelvin Lau

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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:

#### Insertion

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

#### Removal

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

## 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

Per 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 you need to handle a few different scenarios.

#### Case 1: Leaf node

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

#### 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 better 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. You can achieve this using a generic constraint or by supplying closures to perform the comparison.
• 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.
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