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feat: add sql solution to lc problems: No.0608,01965
- No.0608.Tree Node - No.1965.Employees With Missing Information
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‎solution/0600-0699/0608.Tree Node/README.md

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<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>
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<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>
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<li>样例中树的形态如下：
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<p>&nbsp;</p>
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<pre> 1
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<p>&nbsp;</p>
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<pre> 1
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/ \
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2 3
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/ \
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4 5
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</pre>
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<p>&nbsp;</p>
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2 3
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/ \
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4 5
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</pre>
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<p>&nbsp;</p>
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</li>
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</ul>
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<p><strong>注意</strong></p>
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### **SQL**
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```sql
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SELECT id,
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(
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CASE
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WHEN p_id IS NULL THEN 'Root'
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WHEN id IN (
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SELECT p_id
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FROM tree
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) THEN 'Inner'
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ELSE 'Leaf'
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END
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) AS type
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FROM tree;
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```
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<!-- tabs:end -->

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

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

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

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SELECT id,
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(
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CASE
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WHEN p_id IS NULL THEN 'Root'
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WHEN id IN (
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SELECT p_id
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FROM tree
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) THEN 'Inner'
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ELSE 'Leaf'
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END
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) AS type
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FROM tree;

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

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

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

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

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

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SELECT employee_id
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FROM Employees AS e
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WHERE e.employee_id NOT IN (
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SELECT employee_id
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FROM Salaries
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)
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UNION
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SELECT employee_id
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FROM Salaries AS s
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WHERE s.employee_id NOT IN (
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SELECT employee_id
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FROM Employees
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)
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ORDER BY employee_id;

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