Trees

BST Search

Search a binary search tree for one present and one absent value.

Algorithm

The canonical tree is 4(2(1,3),6(5,7)), so this Swift DSA implementation can be compared directly with the rest of the DSA track.

Basic Implementation

basic.swift 
final class Node {
    let value: Int
    var left: Node?
    var right: Node?
    init(_ value: Int, _ left: Node? = nil, _ right: Node? = nil) { self.value = value; self.left = left; self.right = right }
}
func render(_ node: Node?) -> String {
    guard let node = node else { return "_" }
    if node.left == nil && node.right == nil { return String(node.value) }
    return "\(node.value)(\(render(node.left)),\(render(node.right)))"
}
func sampleTree() -> Node {
    return Node(4, Node(2, Node(1), Node(3)), Node(6, Node(5), Node(7)))
}
func listString(_ values: [Int]) -> String { return "[" + values.map(String.init).joined(separator: ", ") + "]" }
func search(_ root: Node?, _ target: Int) -> Bool { var node = root; while let current = node { if target == current.value { return true }; node = target < current.value ? current.left : current.right }; return false }
let root = sampleTree()
print(search(root, 5) ? "5 found" : "5 not found")
print(search(root, 8) ? "8 found" : "8 not found")

Complexity

  • Time: O(h) per search
  • Space: O(1) iterative

Implementation notes

  • Render tree structure explicitly instead of printing node objects.
  • The replay highlights the node, traversal state, queue, path, or search cursor that changes at each step.
search path A comparison chooses one subtree at each step, so whole branches are skipped.