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# Binary Tree

A binary tree is a tree where each node has 0, 1, or 2 children. This is a binary tree:

The child nodes are usually called the left child and the right child. If a node doesn't have any children, it's called a leaf node. The root is the node at the very top of the tree (programmers like their trees upside down).

Often nodes will have a link back to their parent but this is not strictly necessary.

Binary trees are often used as binary search trees. In that case, the nodes must be in a specific order (smaller values on the left, larger values on the right). But this is not a requirement for all binary trees.

For example, here is a binary tree that represents a sequence of arithmetical operations, `(5 * (a - 10)) + (-4 * (3 / b))`:

## The code

Here's how you could implement a general-purpose binary tree in Swift:

```public indirect enum BinaryTree<T> {
case node(BinaryTree<T>, T, BinaryTree<T>)
case empty
}```

As an example of how to use this, let's build that tree of arithmetic operations:

```// leaf nodes
let node5 = BinaryTree.node(.empty, "5", .empty)
let nodeA = BinaryTree.node(.empty, "a", .empty)
let node10 = BinaryTree.node(.empty, "10", .empty)
let node4 = BinaryTree.node(.empty, "4", .empty)
let node3 = BinaryTree.node(.empty, "3", .empty)
let nodeB = BinaryTree.node(.empty, "b", .empty)

// intermediate nodes on the left
let Aminus10 = BinaryTree.node(nodeA, "-", node10)
let timesLeft = BinaryTree.node(node5, "*", Aminus10)

// intermediate nodes on the right
let minus4 = BinaryTree.node(.empty, "-", node4)
let divide3andB = BinaryTree.node(node3, "/", nodeB)
let timesRight = BinaryTree.node(minus4, "*", divide3andB)

// root node
let tree = BinaryTree.node(timesLeft, "+", timesRight)```

You need to build up the tree in reverse, starting with the leaf nodes and working your way up to the top.

It will be useful to add a `description` method so you can print the tree:

```extension BinaryTree: CustomStringConvertible {
public var description: String {
switch self {
case let .node(left, value, right):
return "value: \(value), left = [\(left.description)], right = [\(right.description)]"
case .empty:
return ""
}
}
}```

If you `print(tree)` you should see something like this:

``````value: +, left = [value: *, left = [value: 5, left = [], right = []], right = [value: -, left = [value: a, left = [], right = []], right = [value: 10, left = [], right = []]]], right = [value: *, left = [value: -, left = [], right = [value: 4, left = [], right = []]], right = [value: /, left = [value: 3, left = [], right = []], right = [value: b, left = [], right = []]]]
``````

With a bit of imagination, you can see the tree structure. ;-) It helps if you indent it:

``````value: +,
left = [value: *,
left = [value: 5, left = [], right = []],
right = [value: -,
left = [value: a, left = [], right = []],
right = [value: 10, left = [], right = []]]],
right = [value: *,
left = [value: -,
left = [],
right = [value: 4, left = [], right = []]],
right = [value: /,
left = [value: 3, left = [], right = []],
right = [value: b, left = [], right = []]]]
``````

Another useful method is counting the number of nodes in the tree:

```  public var count: Int {
switch self {
case let .node(left, _, right):
return left.count + 1 + right.count
case .empty:
return 0
}
}```

On the tree from the example, `tree.count` should be 12.

Something you often need to do with trees is traverse them, i.e. look at all the nodes in some order. There are three ways to traverse a binary tree:

1. In-order (or depth-first): first look at the left child of a node, then at the node itself, and finally at its right child.
2. Pre-order: first look at a node, then at its left and right children.
3. Post-order: first look at the left and right children and process the node itself last.

Here is how you'd implement that:

```  public func traverseInOrder(process: (T) -> Void) {
if case let .node(left, value, right) = self {
left.traverseInOrder(process: process)
process(value)
right.traverseInOrder(process: process)
}
}

public func traversePreOrder(process: (T) -> Void) {
if case let .node(left, value, right) = self {
process(value)
left.traversePreOrder(process: process)
right.traversePreOrder(process: process)
}
}

public func traversePostOrder(process: (T) -> Void) {
if case let .node(left, value, right) = self {
left.traversePostOrder(process: process)
right.traversePostOrder(process: process)
process(value)
}
}```

As is common when working with tree structures, these functions call themselves recursively.

For example, if you traverse the tree of arithmetic operations in post-order, you'll see the values in this order:

``````5
a
10
-
*
4
-
3
b
/
*
+
``````

The leaves appear first. The root node appears last.

You can use a stack machine to evaluate these expressions, something like the following pseudocode:

```tree.traversePostOrder { s in
switch s {
case this is a numeric literal, such as 5:
push it onto the stack
case this is a variable name, such as a:
look up the value of a and push it onto the stack
case this is an operator, such as *:
pop the two top-most items off the stack, multiply them,
and push the result back onto the stack
}
the result is in the top-most item on the stack
}```

Written for Swift Algorithm Club by Matthijs Hollemans