A function which, when given a root node, will determine if the tree is symmetric. AKA Breadth-first search

symmetric_tree.rb

#tell whether the tree is visually symmetric (it is)
class Node
  attr_reader :value, :left, :right
  def initialize(value, left, right)
    @value, @left, @right = value, left, right
  end
  
  #added in to see output
  def to_s; @value.to_s; end
  def inspect; @value.to_s; end
end

i = Node.new('i', nil, nil)
g = Node.new('g', nil, i)
f = Node.new('f', nil, nil)
c = Node.new('c', f, g)
e = Node.new('e', nil, nil)
h = Node.new('h', nil, nil)
d = Node.new('d', h, nil)
b = Node.new('b', d, e)
root = Node.new('a', b, c)

def is_tree_symmetric?(node)
  # the queue will be each level of the tree
  queue = [node]
  while !queue.empty?
    # we compare each element with it's mirror on the other side of the queue
    for index in 0..queue.size/2 do
      # we only care if both are nil or both are not nil
      if (queue[index].nil? && queue[queue.size-1-index].nil?) || 
         (!queue[index].nil? && !queue[queue.size-1-index].nil?)
         # expand each element into its children. The element is removed
         queue.each_index do |x|
            sitin = []
            sitin = [queue[x].left] unless queue[x].nil?
            sitin.push(queue[x].right) unless queue[x].nil?
            queue[x] = sitin
          end
          queue = queue.flatten
          # puts queue.inspect
      else
        # if there is anything off, the tree is not symmetric
        return false
      end
    end
    # if the queue is all nils, then we've reached the bottom of the tree
    # compact removes nils, and an empty array means the whole queue was nils
    if queue.compact.empty?
      return true
    end
  end
end

puts is_tree_symmetric?(root)