Btrfs: return gracefully from balance if fs tree is corrupted

When relocating tree blocks, we firstly get block information from
back references in the extent tree, we then search fs tree to try to
find all parents of a block.

However, if fs tree is corrupted, eg. if there're some missing
items, we could come across these WARN_ONs and BUG_ONs.

This makes us print some error messages and return gracefully
from balance.
Signed-off-by: Liu Bo <bo.li.liu@oracle.com>
Reviewed-by: Josef Bacik <jbacik@fb.com>
Signed-off-by: David Sterba <dsterba@suse.com>

fs/btrfs/relocation.c

@@ -923,9 +923,16 @@ struct backref_node *build_backref_tree(struct reloc_control *rc,
 			path2->slots[level]--;
 
 		eb = path2->nodes[level];
-		WARN_ON(btrfs_node_blockptr(eb, path2->slots[level]) !=
-			cur->bytenr);
-
+		if (btrfs_node_blockptr(eb, path2->slots[level]) !=
+		    cur->bytenr) {
+			btrfs_err(root->fs_info,
+	"couldn't find block (%llu) (level %d) in tree (%llu) with key (%llu %u %llu)",
+				  cur->bytenr, level - 1, root->objectid,
+				  node_key->objectid, node_key->type,
+				  node_key->offset);
+			err = -ENOENT;
+			goto out;
+		}
 		lower = cur;
 		need_check = true;
 		for (; level < BTRFS_MAX_LEVEL; level++) {
@@ -2686,11 +2693,15 @@ static int do_relocation(struct btrfs_trans_handle *trans,
 
 		if (!upper->eb) {
 			ret = btrfs_search_slot(trans, root, key, path, 0, 1);
-			if (ret < 0) {
+			if (ret) {
+				if (ret < 0)
 					err = ret;
+				else
+					err = -ENOENT;
+
+				btrfs_release_path(path);
 				break;
 			}
-			BUG_ON(ret > 0);
 
 			if (!upper->eb) {
 				upper->eb = path->nodes[upper->level];