diff --git a/solutions/22. Generate Parentheses/README.md b/solutions/22. Generate Parentheses/README.md
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+---
+comments: true
+difficulty: medium
+# Follow `Topics` tags
+tags:
+    - String
+    - Dynamic Programming
+    - Backtracking
+---
+
+# [22. Generate Parentheses](https://leetcode.com/problems/generate-parentheses/description/)
+
+## Description
+
+Given `n` pairs of parentheses, write a function that returns all possible valid combinations of well-formed (i.e., properly opened and closed) parentheses.
+
+
+**Example 1:**
+```
+Input: n = 2
+Output: ["(())", "()()"]
+```
+
+**Example 2:**
+```
+Input: n = 4
+Output: ["(((())))","((()()))","((())())","((()))()","(()(()))","(()()())","(()())()","(())(())","(())()()","()((()))","()(()())","()(())()","()()(())","()()()()"]
+```
+
+**Constraints:**
+
+* `1 <= n <= 8`
+
+## Solution
+
+Require open parentheses before close parentheses. Therefore need to increase number of open parentheses until n at first, then increase number of close parentheses until n.
+
+```java
+class Solution {
+    public List<String> generateParenthesis(int n) {
+        List<String> result = new ArrayList<>();
+        dfs(result, 0, 0, "", n);
+        return result;
+    }
+
+    public void dfs(List<String> result, int left, int right, String current, int n) {
+        if (current.length() == n * 2) {
+            result.add(current);
+            return;
+        }
+        if (left < n) {
+            dfs(result, left + 1, right, current + "(", n);
+        }
+        if (right < left) {
+            dfs(result, left, right + 1, current + ")", n);
+        }
+    }
+}
+```
+
+```python
+class Solution:
+    def generateParenthesis(self, n: int) -> list[str]:
+        result = []
+        self.dfs(result, 0, 0, "", n)
+        return result
+    
+    def dfs(self, result: list[str], left: int, right: int, current: str, n: int):
+        if left == n and right == n:
+            result.append(current)
+            return
+        if left < n:
+            self.dfs(result, left + 1, right, current + "(", n)
+        if right < left:
+            self.dfs(result, left, right + 1, current + ")", n)
+```
+
+## Complexity
+
+- Time complexity: $$O(2^n)$$
+<!-- Add time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(n)$$
+<!-- Add space complexity here, e.g. $$O(n)$$ -->
diff --git a/solutions/22. Generate Parentheses/Solution.java b/solutions/22. Generate Parentheses/Solution.java
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+import java.util.*;
+
+class Solution {
+    public List<String> generateParenthesis(int n) {
+        List<String> result = new ArrayList<>();
+        dfs(result, 0, 0, "", n);
+        return result;
+    }
+
+    public void dfs(List<String> result, int left, int right, String current, int n) {
+        if (current.length() == n * 2) {
+            result.add(current);
+            return;
+        }
+        if (left < n) {
+            dfs(result, left + 1, right, current + "(", n);
+        }
+        if (right < left) {
+            dfs(result, left, right + 1, current + ")", n);
+        }
+    }
+}
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diff --git a/solutions/22. Generate Parentheses/Solution.py b/solutions/22. Generate Parentheses/Solution.py
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+class Solution:
+    def generateParenthesis(self, n: int) -> list[str]:
+        result = []
+        self.dfs(result, 0, 0, "", n)
+        return result
+    
+    def dfs(self, result: list[str], left: int, right: int, current: str, n: int):
+        if left == n and right == n:
+            result.append(current)
+            return
+        if left < n:
+            self.dfs(result, left + 1, right, current + "(", n)
+        if right < left:
+            self.dfs(result, left, right + 1, current + ")", n)