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

Basic Implementation

basic.rb 
class Node
  attr_accessor :value, :left, :right
  def initialize(value, left = nil, right = nil)
    @value = value
    @left = left
    @right = right
  end
end
def render(node)
  return "_" if node.nil?
  return node.value.to_s if node.left.nil? && node.right.nil?
  "#{node.value}(#{render(node.left)},#{render(node.right)})"
end
def sample_tree
  Node.new(4, Node.new(2, Node.new(1), Node.new(3)), Node.new(6, Node.new(5), Node.new(7)))
end
def search(root, target); node = root; until node.nil?; return true if target == node.value; node = target < node.value ? node.left : node.right; end; false; end
root = sample_tree
puts(search(root, 5) ? '5 found' : '5 not found')
puts(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.