old_contents: """
  Given a binary tree, check whether it is a mirror of
  itself (ie, symmetric around its center).
  
  For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
  
      1
     / \
    2   2
   / \ / \
  3  4 4  3
  But the following [1,2,2,null,3,null,3] is not:
      1
     / \
    2   2
     \   \
     3    3
  Note:
  Bonus points if you could solve it both recursively and iteratively.
  """
  
  # TC: O(b) SC: O(log n)
  def is_symmetric(root):
      if not root:
          return True
      return helper(root.left, root.right)
  
  
  def helper(p, q):
      if not p and not q:
          return True
      if not p or not q or q.val != p.val:
          return False
      return helper(p.left, q.right) and helper(p.right, q.left)
  
  
  def is_symmetric_iterative(root):
      if not root:
          return True
      stack = [[root.left, root.right]]
      while stack:
          left, right = stack.pop()  # popleft
          if not left and not right:
              continue
          if not left or not right:
              return False
          if left.val == right.val:
              stack.append([left.left, right.right])
              stack.append([left.right, right.right])
          else:
              return False
      return True
  
new_contents: """
  Given a binary tree, check whether it is a mirror of
  itself (ie, symmetric around its center).
  
  For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
  
      1
     / \
    2   2
   / \ / \
  3  4 4  3
  But the following [1,2,2,null,3,null,3] is not:
      1
     / \
    2   2
     \   \
     3    3
  Note:
  Bonus points if you could solve it both recursively and iteratively.
  """
  
  # TC: O(b) SC: O(log n)
  def is_symmetric(root):
      if not root:
          return True
      return helper(root.left, root.right)
  
  
  def helper(p, q):
      if not p and not q:
          return True
      if not p or not q or q.val != p.val:
          return False
      return helper(p.left, q.right) and helper(p.right, q.left)
  
  
  def is_symmetric_iterative(root):
      if not root:
          return True
      stack = [[root.left, root.right]]
      while stack:
          left, right = stack.pop()  # popleft
          if not left and not right:
              continue
          if not left or not right:
              return False
          if left.val == right.val:
              stack.append([left.left, right.right])
              stack.append([left.right, right.left])
          else:
              return False
      return True
  
subject: Update is_symmetric.py (#396)