Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level)..

level_order.rb

# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} root
# @return {Integer[][]}
def level_order(root)
  nodes = [root].compact
  res   = []

  loop do
    nodes_size = nodes.size
    break if nodes_size.zero?

    curr_res = []

    while nodes_size > 0 do
      curr = nodes.shift
      nodes_size -= 1

      curr_res << curr.val

      nodes << curr.left if curr.left
      nodes << curr.right if curr.right
    end

    res << curr_res
  end

  res
end

level_order_rec.rb

# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} root
# @return {Integer[][]}

def level_order(root)
  res = []
  dfs(root, res)
  res
end

def dfs(node, res, level = 0)
  return if node.nil?

  res[level] ||= []
  res[level] << node.val

  dfs(node.left, res, level + 1)
  dfs(node.right, res, level + 1)
end