feat: add sql solution to lc problems: No.0608,01965

- No.0608.Tree Node
- No.1965.Employees With Missing Information

Author: YangFong

solution/0600-0699/0608.Tree Node/README.md

@@ -52,19 +52,15 @@
 	<li>节点 &#39;2&#39; 是内部节点，因为它有父节点 &#39;1&#39; ，也有孩子节点 &#39;4&#39; 和 &#39;5&#39; 。</li>
 	<li>节点 &#39;3&#39;, &#39;4&#39; 和 &#39;5&#39; 都是叶子节点，因为它们都有父节点同时没有孩子节点。</li>
 	<li>样例中树的形态如下：
-	<p>&nbsp;</p>
-
-    <pre>			  1
+		<p>&nbsp;</p>
+    	<pre>			  1
     		/   \
-                      2       3
-                    /   \
-                  4       5
-
-</pre>
-
-    <p>&nbsp;</p>
+          2       3
+		/   \
+	   4     5
+		</pre>
+    	<p>&nbsp;</p>
     </li>
-
 </ul>
 
 <p><strong>注意</strong></p>
@@ -80,7 +76,18 @@
 ### **SQL**
 
 ```sql
-
+SELECT id,
+    (
+        CASE
+            WHEN p_id IS NULL THEN 'Root'
+            WHEN id IN (
+                SELECT p_id
+                FROM tree
+            ) THEN 'Inner'
+            ELSE 'Leaf'
+        END
+    ) AS type
+FROM tree;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0608.Tree Node/README_EN.md

@@ -91,7 +91,18 @@ Tree table:
 ### **SQL**
 
 ```sql
-
+SELECT id,
+    (
+        CASE
+            WHEN p_id IS NULL THEN 'Root'
+            WHEN id IN (
+                SELECT p_id
+                FROM tree
+            ) THEN 'Inner'
+            ELSE 'Leaf'
+        END
+    ) AS type
+FROM tree;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0608.Tree Node/Solution.sql

@@ -0,0 +1,12 @@
+SELECT id,
+    (
+        CASE
+            WHEN p_id IS NULL THEN 'Root'
+            WHEN id IN (
+                SELECT p_id
+                FROM tree
+            ) THEN 'Inner'
+            ELSE 'Leaf'
+        END
+    ) AS type
+FROM tree;

solution/1900-1999/1965.Employees With Missing Information/README.md

@@ -82,7 +82,20 @@ Result table:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+SELECT employee_id
+FROM Employees AS e
+WHERE e.employee_id NOT IN (
+        SELECT employee_id
+        FROM Salaries
+    )
+UNION
+SELECT employee_id
+FROM Salaries AS s
+WHERE s.employee_id NOT IN (
+        SELECT employee_id
+        FROM Employees
+    )
+ORDER BY employee_id;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1965.Employees With Missing Information/README_EN.md

@@ -86,7 +86,20 @@ The salary of employee 2 is missing.
 ### **SQL**
 
 ```sql
-
+SELECT employee_id
+FROM Employees AS e
+WHERE e.employee_id NOT IN (
+        SELECT employee_id
+        FROM Salaries
+    )
+UNION
+SELECT employee_id
+FROM Salaries AS s
+WHERE s.employee_id NOT IN (
+        SELECT employee_id
+        FROM Employees
+    )
+ORDER BY employee_id;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1965.Employees With Missing Information/Solution.sql

@@ -0,0 +1,14 @@
+SELECT employee_id
+FROM Employees AS e
+WHERE e.employee_id NOT IN (
+        SELECT employee_id
+        FROM Salaries
+    )
+UNION
+SELECT employee_id
+FROM Salaries AS s
+WHERE s.employee_id NOT IN (
+        SELECT employee_id
+        FROM Employees
+    )
+ORDER BY employee_id;