Swift binary tree list of nodes at given depth - swift

Question

I'm writing a Swift algorithm for a binary tree. My goal is to create a list of nodes at specific depth something like
func listNodeAt(_n: Int) --> [T] {
}
Here is my tree class
public class BinaryTreeNode<T:Comparable> {
//Value and children vars
public var value:T
public var leftChild:BinaryTreeNode?
public var rightChild:BinaryTreeNode?
public weak var parent:BinaryTreeNode?
//Initialization
public convenience init(value: T) {
self.init(value: value, left: nil, right: nil, parent:nil)
}
public init(value:T, left:BinaryTreeNode?, right:BinaryTreeNode?, parent:BinaryTreeNode?) {
self.value = value
self.leftChild = left
self.rightChild = right
self.parent = parent
}
}
I have build a helper function to calculate the depth of a Node
//Depth
public func depth() -> Int {
guard var node = parent else {
return 0
}
var depth = 1
while let parent = node.parent {
depth = depth + 1
node = parent
}
return depth
}
How can we achieve the desired function?

Answer 1

func listNodeAt(_ n: Int) -> [T] {
return getElementsAt(n, node: self)
}
private func getElementsAt(_ n: Int, node: BinaryTreeNode<T>, traversingDepth: Int = 0) -> [T] {
var array = Array<T>()
if traversingDepth < n {
if let left = node.leftChild {
array = array + getElementsAt(n, node: left, traversingDepth: traversingDepth + 1)
}
if let right = node.rightChild {
array = array + getElementsAt(n, node: right, traversingDepth: traversingDepth + 1)
}
} else if traversingDepth == n {
array.append(node.value)
}
return array
}
This is one of the solution. Assuming here the self is the root node.