SQL Server 2019

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CREATE TABLE [dbo].[Equipament](
	[ID] [nvarchar](50) NOT NULL,
	[AL] [nvarchar](50) NULL,
	[LVL] [int] NULL,
	[IDParent] [nvarchar](50) NULL,
 CONSTRAINT [PK_Equipament] PRIMARY KEY CLUSTERED 
(
	[ID] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON, OPTIMIZE_FOR_SEQUENTIAL_KEY = OFF) ON [PRIMARY]
) ON [PRIMARY]
ALTER TABLE [dbo].[Equipament]  WITH CHECK ADD  CONSTRAINT [FK_Equipament_Equipament] FOREIGN KEY([IDParent])
REFERENCES [dbo].[Equipament] ([ID])
INSERT INTO [Equipament]
           ([ID]
           ,[AL]
           ,[LVL]
           ,[IDParent])
     VALUES
           ('AA'
			,'07BB'
           ,1
           ,NULL),

		   ('AB'
			,'07BB'
           ,1
           ,NULL),

		   ('BA'
			,'07BB'
           ,2
           ,'AA'),

		   ('BB'
			,'07BB'
           ,2
           ,'AB'),

		   ('BC'
			,'07BB'
           ,2
           ,'AB'),

		   ('CA'
			,'07BB'
           ,3
           ,'BB'),

		   ('CB'
			,'07BB'
           ,3
           ,'BC'),

		   ('CD'
			,'07BB'
           ,3
           ,'BC'),

		   ('CE'
			,'07BB'
           ,3
           ,'AA')
9 rows affected

select * from [Equipament]


ID
AL
LVL
IDParent


AA
07BB
1
null

AB
07BB
1
null

BA
07BB
2
AA

BB
07BB
2
AB

BC
07BB
2
AB

CA
07BB
3
BB

CB
07BB
3
BC

CD
07BB
3
BC

CE
07BB
3
AA


CREATE TABLE [dbo].[Conns](
	[Conn] [nvarchar](50) NOT NULL,
	[IDEquip] [nvarchar](50) NOT NULL,
 CONSTRAINT [PK_Ligacoes] PRIMARY KEY CLUSTERED 
(
	[Conn] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON, OPTIMIZE_FOR_SEQUENTIAL_KEY = OFF) ON [PRIMARY]
) ON [PRIMARY]

ALTER TABLE [dbo].[Conns]  WITH CHECK ADD  CONSTRAINT [FK_Ligacoes_Equipament] FOREIGN KEY([IDEquip])
REFERENCES [dbo].[Equipament] ([ID])
INSERT INTO [dbo].[Conns]
           ([Conn]
           ,[IDEquip]
)
     VALUES
           ('123A'
           ,'CB'
			),

			('123B'
           ,'CB'
			),

			('123C'
           ,'CB'
			),

			('234A'
           ,'CD'
			),

			('234B'
           ,'CE'
			)
5 rows affected

;WITH QUERY AS
(
    SELECT E.ID,E.IDParent,L.conn
    FROM Equipament E
    LEFT JOIN conns L ON E.ID=L.IDEquip
    WHERE E.IDParent IS NULL

    UNION ALL

    SELECT E.ID,E.IDParent,L.conn
    FROM Equipament E
    JOIN conns L ON E.ID = L.IDEquip
    JOIN QUERY P on P.ID = E.IDParent
)
SELECT
    E.ID,
    SUM(S.Total) AS LTotal
FROM Equipament E
LEFT JOIN ( SELECT Q.ID, COUNT(Q.conn) AS Total
                FROM QUERY Q
                GROUP BY Q.ID
            ) as S 
ON E.ID = S.ID
GROUP BY E.ID
ORDER BY E.ID
option (maxrecursion 0)


ID
LTotal


AA
0

AB
0

BA
null

BB
null

BC
null

CA
null

CB
null

CD
null

CE
1


Warning: Null value is eliminated by an aggregate or other SET operation.
ID	AL	LVL	IDParent
AA	07BB	1	null
AB	07BB	1	null
BA	07BB	2	AA
BB	07BB	2	AB
BC	07BB	2	AB
CA	07BB	3	BB
CB	07BB	3	BC
CD	07BB	3	BC
CE	07BB	3	AA