// Edge cases if node is null return null
// If one node without any children, nothing to swap
// If one node exists, swap children and then recursively swap left/right children trees
// O(N) time where N is number of nodes in tree, space is O(N) where the height could be at most the number of nodes
function invert(node) {
  // Base Cases:
  // If no node, return null
  if (!node) {
    return null;
  }
  const tempNode = node.left;
  node.left = node.right;
  node.right = tempNode;
  // If node with children, swap left/right and then recursively invert left/right subtree
  invert(node.left);
  invert(node.right);
  return node;
}