feat: update solutions to lc problems: No.0534,0608,1264,1965 (#1791)

* No.0534.Game Play Analysis III
* No.0608.Tree Node
* No.1264.Page Recommendations
* No.1965.Employees With Missing Information

Author: yanglbme

solution/0500-0599/0534.Game Play Analysis III/README.md

@@ -66,9 +66,13 @@ Activity table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
-**方法一：SUM() OVER() 窗口函数**
+**方法一：使用窗口函数**
 
-我们可以使用 `SUM() OVER()` 窗口函数来计算每个玩家到目前为止玩了多少游戏。在 `OVER()` 子句中，我们使用 `PARTITION BY` 子句将玩家分组，然后使用 `ORDER BY` 子句按日期排序。
+我们可以使用窗口函数 `SUM() OVER()`，按照 `player_id` 分组，按照 `event_date` 排序，计算每个用户截止到当前日期的游戏总数。
+
+**方法二：使用自连接 + 分组**
+
+我们也可以使用自连接，将 `Activity` 表自连接，连接条件为 `t1.player_id = t2.player_id AND t1.event_date >= t2.event_date`，然后按照 `t1.player_id` 和 `t1.event_date` 分组，累计 `t2.games_played`，得到每个用户截止到当前日期的游戏总数。
 
 <!-- tabs:start -->
 
@@ -86,4 +90,29 @@ SELECT
 FROM Activity;
 ```
 
+```sql
+# Write your MySQL query statement below
+SELECT
+    t1.player_id,
+    t1.event_date,
+    sum(t2.games_played) AS games_played_so_far
+FROM
+    Activity AS t1,
+    Activity AS t2
+WHERE t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
+GROUP BY 1, 2;
+```
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    t1.player_id,
+    t1.event_date,
+    sum(t2.games_played) AS games_played_so_far
+FROM
+    Activity AS t1
+    CROSS JOIN Activity AS t2 ON t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
+GROUP BY 1, 2;
+```
+
 <!-- tabs:end -->

solution/0500-0599/0534.Game Play Analysis III/README_EN.md

@@ -61,6 +61,14 @@ Note that for each player we only care about the days when the player logged in.
 
 ## Solutions
 
+**Solution 1: Window Function**
+
+We can use the window function `SUM() OVER()` to group by `player_id`, sort by `event_date`, and calculate the total number of games played by each user up to the current date.
+
+**Solution 2: Self-Join + Group By**
+
+We can also use a self-join to join the `Activity` table with itself on the condition of `t1.player_id = t2.player_id AND t1.event_date >= t2.event_date`, and then group by `t1.player_id` and `t1.event_date`, and calculate the cumulative sum of `t2.games_played`. This will give us the total number of games played by each user up to the current date.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -77,4 +85,29 @@ SELECT
 FROM Activity;
 ```
 
+```sql
+# Write your MySQL query statement below
+SELECT
+    t1.player_id,
+    t1.event_date,
+    sum(t2.games_played) AS games_played_so_far
+FROM
+    Activity AS t1,
+    Activity AS t2
+WHERE t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
+GROUP BY 1, 2;
+```
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    t1.player_id,
+    t1.event_date,
+    sum(t2.games_played) AS games_played_so_far
+FROM
+    Activity AS t1
+    CROSS JOIN Activity AS t2 ON t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
+GROUP BY 1, 2;
+```
+
 <!-- tabs:end -->

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

@@ -91,6 +91,14 @@ Tree table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：条件判断 + 子查询**
+
+我们可以使用 `CASE WHEN` 条件判断语句来判断每个节点的类型，具体地：
+
+-   如果一个节点的 `p_id` 为 `NULL`，则该节点为根节点；
+-   否则，如果一个节点是另一个节点的父节点（这里我们使用子查询来判断），则该节点为内部节点；
+-   否则，该节点为叶子节点。
+
 <!-- tabs:start -->
 
 ### **SQL**

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

@@ -86,6 +86,14 @@ Tree table:
 
 ## Solutions
 
+**Solution 1: Conditional Statements + Subquery**
+
+We can use the `CASE WHEN` conditional statement to determine the type of each node as follows:
+
+-   If a node's `p_id` is `NULL`, then it is a root node.
+-   Otherwise, if a node is the parent node of another node (we use a subquery to determine this), then it is an internal node.
+-   Otherwise, it is a leaf node.
+
 <!-- tabs:start -->
 
 ### **SQL**

solution/1200-1299/1264.Page Recommendations/README.md

@@ -96,10 +96,29 @@ Likes table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：合并 + 等值连接 + 子查询**
+
+我们先查出所有与 `user_id = 1` 的用户是朋友的用户，记录在 `T` 表中，然后再查出所有在 `T` 表中的用户喜欢的页面，最后排除掉 `user_id = 1` 喜欢的页面即可。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT user1_id AS user_id FROM Friendship WHERE user2_id = 1
+        UNION
+        SELECT user2_id AS user_id FROM Friendship WHERE user1_id = 1
+    )
+SELECT DISTINCT page_id AS recommended_page
+FROM
+    T
+    JOIN Likes USING (user_id)
+WHERE page_id NOT IN (SELECT page_id FROM Likes WHERE user_id = 1);
+```
+
 ```sql
 # Write your MySQL query statement below
 SELECT DISTINCT page_id AS recommended_page

solution/1200-1299/1264.Page Recommendations/README_EN.md

@@ -90,10 +90,29 @@ Page 88 is not suggested because user 1 already likes it.
 
 ## Solutions
 
+**Solution 1: Union + Equi-Join + Subquery**
+
+First, we query all users who are friends with `user_id = 1` and record them in the `T` table. Then, we query all pages that users in the `T` table like, and finally exclude the pages that `user_id = 1` likes.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT user1_id AS user_id FROM Friendship WHERE user2_id = 1
+        UNION
+        SELECT user2_id AS user_id FROM Friendship WHERE user1_id = 1
+    )
+SELECT DISTINCT page_id AS recommended_page
+FROM
+    T
+    JOIN Likes USING (user_id)
+WHERE page_id NOT IN (SELECT page_id FROM Likes WHERE user_id = 1);
+```
+
 ```sql
 # Write your MySQL query statement below
 SELECT DISTINCT page_id AS recommended_page

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

@@ -83,29 +83,26 @@ Salaries table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：子查询 + 合并**
+
+我们可以先从 `Employees` 表中找出所有不在 `Salaries` 表中的 `employee_id`，再从 `Salaries` 表中找出所有不在 `Employees` 表中的 `employee_id`，最后将两个结果合并，然后按照 `employee_id` 排序即可。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
+# Write your MySQL query statement below
 SELECT employee_id
-FROM Employees AS e
-WHERE
-    e.employee_id NOT IN (
-        SELECT employee_id
-        FROM Salaries
-    )
+FROM Employees
+WHERE 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;
+FROM Salaries
+WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -81,27 +81,24 @@ The salary of employee 2 is missing.
 
 ## Solutions
 
+**Solution 1: Subquery + Union**
+
+We can first find all `employee_id` that are not in the `Salaries` table from the `Employees` table, and then find all `employee_id` that are not in the `Employees` table from the `Salaries` table. Finally, we can combine the two results using the `UNION` operator, and sort the result by `employee_id`.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT employee_id
-FROM Employees AS e
-WHERE
-    e.employee_id NOT IN (
-        SELECT employee_id
-        FROM Salaries
-    )
+FROM Employees
+WHERE 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;
+FROM Salaries
+WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

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

@@ -1,16 +1,9 @@
+# Write your MySQL query statement below
 SELECT employee_id
-FROM Employees AS e
-WHERE
-    e.employee_id NOT IN (
-        SELECT employee_id
-        FROM Salaries
-    )
+FROM Employees
+WHERE 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;
+FROM Salaries
+WHERE employee_id NOT IN (SELECT employee_id FROM Employees)
+ORDER BY 1;