Btrfs: ensure btrfs_prev_leaf doesn't miss 1 item

		return ret;

	btrfs_item_key(path->nodes[0], &found_key, 0);

	ret = comp_keys(&found_key, &key);

	if (ret < 0)

	/*

	 * We might have had an item with the previous key in the tree right

	 * before we released our path. And after we released our path, that

	 * item might have been pushed to the first slot (0) of the leaf we

	 * were holding due to a tree balance. Alternatively, an item with the

	 * previous key can exist as the only element of a leaf (big fat item).

	 * Therefore account for these 2 cases, so that our callers (like

	 * btrfs_previous_item) don't miss an existing item with a key matching

	 * the previous key we computed above.

	 */

	if (ret <= 0)

		return 0;

	return 1;

}