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 Kotlin DSA implementation can be compared directly with the rest of the DSA track.

Basic Implementation

basic.kt 
class Node(val value: Int, var left: Node? = null, var right: Node? = null)
fun render(node: Node?): String {
    if (node == null) return "_"
    if (node.left == null && node.right == null) return node.value.toString()
    return "${node.value}(${render(node.left)},${render(node.right)})"
}
fun sampleTree() = Node(4, Node(2, Node(1), Node(3)), Node(6, Node(5), Node(7)))
fun listString(values: List<Int>) = values.joinToString(", ", "[", "]")
fun search(root: Node?, target: Int): Boolean { var node = root; while (node != null) { if (target == node.value) return true; node = if (target < node.value) node.left else node.right }; return false }
fun main() { val root = sampleTree(); println(if (search(root, 5)) "5 found" else "5 not found"); println(if (search(root, 8)) "8 found" else "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.