int btrfs_prev_leaf(struct btrfs_root *root, struct btrfs_path *path)
{
	struct btrfs_key key;
	struct btrfs_disk_key found_key;
	int ret;

	btrfs_item_key_to_cpu(path->nodes[0], &key, 0);

	if (key.offset > 0) {
		key.offset--;
	} else if (key.type > 0) {
		key.type--;
		key.offset = (u64)-1;
	} else if (key.objectid > 0) {
		key.objectid--;
		key.type = (u8)-1;
		key.offset = (u64)-1;
	} else {
		return 1;
	}

	btrfs_release_path(path);
	ret = btrfs_search_slot(NULL, root, &key, path, 0, 0);
	if (ret < 0)
		return ret;
	btrfs_item_key(path->nodes[0], &found_key, 0);
	ret = comp_keys(&found_key, &key);
	/*
	 * 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;
}
int btrfs_prev_leaf(struct btrfs_root *root, struct btrfs_path *path)
{
	struct btrfs_key key;
	struct btrfs_disk_key found_key;
	int ret;

	btrfs_item_key_to_cpu(path->nodes[0], &key, 0);

	if (key.offset > 0) {
		key.offset--;
	} else if (key.type > 0) {
		key.type--;
		key.offset = (u64)-1;
	} else if (key.objectid > 0) {
		key.objectid--;
		key.type = (u8)-1;
		key.offset = (u64)-1;
	} else {
		return 1;
	}

	btrfs_release_path(path);
	ret = btrfs_search_slot(NULL, root, &key, path, 0, 0);
	if (ret < 0)
		return ret;
	btrfs_item_key(path->nodes[0], &found_key, 0);
	ret = comp_keys(&found_key, &key);
	 
	if (ret <= 0)
		return 0;
	return 1;
}