Classificação com base nos registros próximos e anteriores no SQL

1. Identifique a primeira e a última letra de cada pai usando ROW_NUMBER() e PARTITION BY
2. Corresponder ao último registro do id anterior com o primeiro registro do próximo id .
3. Verifique se o segundo ID pai tem alguma letra que corresponda à letra selecionada acima
4. Use um LEFT JOIN e use CASE ou ISNULL para definir uma prioridade mais alta para tal id registro em que a letra correspondia

;WITH CTE AS 
(
SELECT id,ParentID,letter,
ROW_NUMBER()OVER(PARTITION BY parentId ORDER BY ID) first_element,
ROW_NUMBER()OVER(PARTITION BY parentId ORDER BY ID DESC) Last_element
FROM Child
), CTE2 AS 
(
SELECT c1.id,c1.parentid,c1.letter,c2.parentid as c2parentid
FROM CTE c1
INNER JOIN CTE c2
ON c1.last_element = 1
AND c2.first_element = 1
AND c1.id +1 = c2.id
), CTE3 AS 
(
SELECT C.parentid,C.id
FROM CTE2
INNER JOIN child C ON CTE2.c2parentid = C.parentid
AND C.letter = CTE2.letter
)
SELECT P.number, C.letter
FROM Child C
JOIN Parent P ON C.parentId = P.id
LEFT JOIN CTE3 ON CTE3.id = C.id
ORDER BY P.number, ISNULL(CTE3.id,0) DESC, C.letter 

number  letter
1   A
1   C
2   C
2   B
3   B
3   D

;WITH CTE AS 
(
SELECT id,ParentID,letter,
ROW_NUMBER()OVER(ORDER BY ID) seq_id,
ROW_NUMBER()OVER(PARTITION BY parentId ORDER BY ID) first_element,
ROW_NUMBER()OVER(PARTITION BY parentId ORDER BY ID DESC) Last_element
FROM Child
), CTE2 AS 
(
SELECT c1.id,c1.parentid,c1.letter,c2.parentid as c2parentid
FROM CTE c1
INNER JOIN CTE c2
ON c1.last_element = 1
AND c2.first_element = 1
AND c1.seq_id + 1 = c2.seq_id
)