Level-Order Traversal

Visit a tree breadth-first with a queue.

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
queue = [sample_tree]
output = []
until queue.empty?
  node = queue.shift
  output << node.value
  queue << node.left if node.left
  queue << node.right if node.right
end
puts "[#{output.join(', ')}]"

Complexity

Time: O(n)
Space: O(w) queue space

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.

level order Level-order traversal uses a queue to visit shallower nodes first.