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: Ob SC: Olog n
# TC: O(b) SC: O(log n)
def is_symmetric(root):
    if root is None:
        return True
    return helper(root.left, root.right)


def helper(p, q):
    if p is None and q is None:
        return True
    if p is not None or q is not None 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 root is None:
        return True
    stack = [[root.left, root.right]]
    while stack:
        left, right = stack.pop()  # popleft
        if left is None and right is None:
            continue
        if left is None or right is None:
            return False
        if left.val == right.val:
            stack.append([left.left, right.right])
            stack.append([left.right, right.left])
        else:
            return False
    return True
