diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md
index 063b5429..31c24fc5 100644
--- a/problems/0101.对称二叉树.md
+++ b/problems/0101.对称二叉树.md
@@ -224,8 +224,8 @@ public:
         st.push(root->left);
         st.push(root->right);
         while (!st.empty()) {
-            TreeNode* leftNode = st.top(); st.pop();
             TreeNode* rightNode = st.top(); st.pop();
+            TreeNode* leftNode = st.top(); st.pop();
             if (!leftNode && !rightNode) {
                 continue;
             }
@@ -950,3 +950,4 @@ public bool IsSymmetric(TreeNode root)
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>
 </a>
+
diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md
index ac03ddbb..61488f03 100644
--- a/problems/0242.有效的字母异位词.md
+++ b/problems/0242.有效的字母异位词.md
@@ -383,6 +383,31 @@ object Solution {
     }
 ```
 
+### C
+
+```c
+bool isAnagram(char* s, char* t) {
+    int len1 = strlen(s), len2 = strlen(t);
+    if (len1 != len2) {
+        return false;
+    }
+
+    int map1[26] = {0}, map2[26] = {0};
+    for (int i = 0; i < len1; i++) {
+        map1[s[i] - 'a'] += 1;
+        map2[t[i] - 'a'] += 1;
+    }
+    
+    for (int i = 0; i < 26; i++) {
+        if (map1[i] != map2[i]) {
+            return false;
+        }
+    }
+    
+    return true;
+}
+```
+
 ## 相关题目
 
 * [383.赎金信](https://programmercarl.com/0383.%E8%B5%8E%E9%87%91%E4%BF%A1.html)
diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md
index 83dea97e..6231c22b 100644
--- a/problems/0454.四数相加II.md
+++ b/problems/0454.四数相加II.md
@@ -185,22 +185,28 @@ class Solution:
 ### Go：
 
 ```go
-func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int {
-    m := make(map[int]int)   //key:a+b的数值，value:a+b数值出现的次数
-    count := 0
-     // 遍历nums1和nums2数组，统计两个数组元素之和，和出现的次数，放到map中
+func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int {  
+    m := make(map[int]int)  
+    count := 0  
+  
+    // 构建nums1和nums2的和的map  
     for _, v1 := range nums1 {  
-        for _, v2 := range nums2 {
-            m[v1+v2]++
-        }
-    }
-    // 遍历nums3和nums4数组，找到如果 0-(c+d) 在map中出现过的话，就把map中key对应的value也就是出现次数统计出来
-    for _, v3 := range nums3 {
-        for _, v4 := range nums4 {
-            count += m[-v3-v4]
-        }
-    }
-    return count
+        for _, v2 := range nums2 {  
+            m[v1+v2]++  
+        }  
+    }  
+  
+    // 遍历nums3和nums4，检查-c-d是否在map中  
+    for _, v3 := range nums3 {  
+        for _, v4 := range nums4 {  
+            sum := -v3 - v4  
+            if countVal, ok := m[sum]; ok {  
+                count += countVal  
+            }  
+        }  
+    }  
+  
+    return count  
 }
 ```
 
diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md
index 5604cd56..d86146d6 100644
--- a/problems/0704.二分查找.md
+++ b/problems/0704.二分查找.md
@@ -174,13 +174,17 @@ class Solution {
         int left = 0, right = nums.length - 1;
         while (left <= right) {
             int mid = left + ((right - left) >> 1);
-            if (nums[mid] == target)
+            if (nums[mid] == target) {
                 return mid;
-            else if (nums[mid] < target)
+            }
+            else if (nums[mid] < target) {
                 left = mid + 1;
-            else if (nums[mid] > target)
+            }
+            else { // nums[mid] > target
                 right = mid - 1;
+            }
         }
+        // 未找到目标值
         return -1;
     }
 }
@@ -194,13 +198,17 @@ class Solution {
         int left = 0, right = nums.length;
         while (left < right) {
             int mid = left + ((right - left) >> 1);
-            if (nums[mid] == target)
+            if (nums[mid] == target) {
                 return mid;
-            else if (nums[mid] < target)
+            }
+            else if (nums[mid] < target) {
                 left = mid + 1;
-            else if (nums[mid] > target)
+            }
+            else { // nums[mid] > target
                 right = mid;
+            }
         }
+        // 未找到目标值
         return -1;
     }
 }
diff --git a/problems/1020.飞地的数量.md b/problems/1020.飞地的数量.md
index ff59411e..f708e4a3 100644
--- a/problems/1020.飞地的数量.md
+++ b/problems/1020.飞地的数量.md
@@ -605,6 +605,63 @@ func bfs(grid [][]int, i, j int) {
 }
 ```
 
+### JavaScript
+
+```js
+/**
+ * @param {number[][]} grid
+ * @return {number}
+ */
+var numEnclaves = function (grid) {
+  let row = grid.length;
+  let col = grid[0].length;
+  let count = 0;
+
+  // Check the first and last row, if there is a 1, then change all the connected 1s to 0 and don't count them.
+  for (let j = 0; j < col; j++) {
+    if (grid[0][j] === 1) {
+      dfs(0, j, false);
+    }
+    if (grid[row - 1][j] === 1) {
+      dfs(row - 1, j, false);
+    }
+  }
+
+  // Check the first and last column, if there is a 1, then change all the connected 1s to 0 and don't count them.
+  for (let i = 0; i < row; i++) {
+    if (grid[i][0] === 1) {
+      dfs(i, 0, false);
+    }
+    if (grid[i][col - 1] === 1) {
+      dfs(i, col - 1, false);
+    }
+  }
+
+  // Check the rest of the grid, if there is a 1, then change all the connected 1s to 0 and count them.
+  for (let i = 1; i < row - 1; i++) {
+    for (let j = 1; j < col - 1; j++) {
+      dfs(i, j, true);
+    }
+  }
+
+  function dfs(i, j, isCounting) {
+    let condition = i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] === 0;
+
+    if (condition) return;
+    if (isCounting) count++;
+
+    grid[i][j] = 0;
+
+    dfs(i - 1, j, isCounting);
+    dfs(i + 1, j, isCounting);
+    dfs(i, j - 1, isCounting);
+    dfs(i, j + 1, isCounting);
+  }
+
+  return count;
+};
+```
+
 ### Rust
 
 dfs:
@@ -700,3 +757,4 @@ impl Solution {
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">
   <img src="../pics/网站星球宣传海报.jpg" width="1000"/>
 </a>
+
diff --git a/problems/1207.独一无二的出现次数.md b/problems/1207.独一无二的出现次数.md
index 5c5f92c3..cd89522e 100644
--- a/problems/1207.独一无二的出现次数.md
+++ b/problems/1207.独一无二的出现次数.md
@@ -40,9 +40,9 @@
 回归本题，**本题强调了-1000 <= arr[i] <= 1000**，那么就可以用数组来做哈希，arr[i]作为哈希表（数组）的下标，那么arr[i]可以是负数，怎么办？负数不能做数组下标。
 
 
-**此时可以定义一个2000大小的数组，例如int count[2002];**，统计的时候，将arr[i]统一加1000，这样就可以统计arr[i]的出现频率了。
+**此时可以定义一个2001大小的数组，例如int count[2001];**，统计的时候，将arr[i]统一加1000，这样就可以统计arr[i]的出现频率了。
 
-题目中要求的是是否有相同的频率出现，那么需要再定义一个哈希表（数组）用来记录频率是否重复出现过，bool fre[1002]; 定义布尔类型的就可以了，**因为题目中强调1 <= arr.length <= 1000，所以哈希表大小为1000就可以了**。
+题目中要求的是是否有相同的频率出现，那么需要再定义一个哈希表（数组）用来记录频率是否重复出现过，bool fre[1001]; 定义布尔类型的就可以了，**因为题目中强调1 <= arr.length <= 1000，所以哈希表大小为1000就可以了**。
 
 如图所示：
 
@@ -55,11 +55,11 @@ C++代码如下：
 class Solution {
 public:
     bool uniqueOccurrences(vector<int>& arr) {
-        int count[2002] = {0}; // 统计数字出现的频率
+        int count[2001] = {0}; // 统计数字出现的频率
         for (int i = 0; i < arr.size(); i++) {
             count[arr[i] + 1000]++;
         }
-        bool fre[1002] = {false}; // 看相同频率是否重复出现
+        bool fre[1001] = {false}; // 看相同频率是否重复出现
         for (int i = 0; i <= 2000; i++) {
             if (count[i]) {
                 if (fre[count[i]] == false) fre[count[i]] = true;
@@ -78,7 +78,7 @@ public:
 ```java
 class Solution {
     public boolean uniqueOccurrences(int[] arr) {
-        int[] count = new int[2002];
+        int[] count = new int[2001];
         for (int i = 0; i < arr.length; i++) {
             count[arr[i] + 1000]++; // 防止负数作为下标
         }
@@ -103,10 +103,10 @@ class Solution {
 # 方法 1: 数组在哈西法的应用
 class Solution:
     def uniqueOccurrences(self, arr: List[int]) -> bool:
-        count = [0] * 2002
+        count = [0] * 2001
         for i in range(len(arr)):
             count[arr[i] + 1000] += 1 # 防止负数作为下标
-        freq = [False] * 1002 # 标记相同频率是否重复出现
+        freq = [False] * 1001 # 标记相同频率是否重复出现
         for i in range(2001):
             if count[i] > 0:
                 if freq[count[i]] == False:
@@ -139,12 +139,12 @@ class Solution:
 ``` javascript
 // 方法一：使用数组记录元素出现次数
 var uniqueOccurrences = function(arr) {
-    const count = new Array(2002).fill(0);// -1000 <= arr[i] <= 1000
+    const count = new Array(2001).fill(0);// -1000 <= arr[i] <= 1000
     for(let i = 0; i < arr.length; i++){
         count[arr[i] + 1000]++;// 防止负数作为下标
     }
     // 标记相同频率是否重复出现
-    const fre = new Array(1002).fill(false);// 1 <= arr.length <= 1000
+    const fre = new Array(1001).fill(false);// 1 <= arr.length <= 1000
     for(let i = 0; i <= 2000; i++){
         if(count[i] > 0){//有i出现过
             if(fre[count[i]] === false) fre[count[i]] = true;//之前未出现过，标记为出现
diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md
index a1d49e98..f2a97f4d 100644
--- a/problems/二叉树的递归遍历.md
+++ b/problems/二叉树的递归遍历.md
@@ -671,6 +671,62 @@ public void Traversal(TreeNode cur, IList<int> res)
 }
 ```
 
+### PHP
+```php
+// 144.前序遍历
+function preorderTraversal($root) {
+    $output = [];
+    $this->traversal($root, $output);
+    return $output;
+}
+
+function traversal($root, array &$output) {
+    if ($root->val === null) {
+        return;
+    }
+
+    $output[] = $root->val;
+    $this->traversal($root->left, $output);
+    $this->traversal($root->right, $output);
+}
+```
+```php
+// 94.中序遍历
+function inorderTraversal($root) {
+    $output = [];
+    $this->traversal($root, $output);
+    return $output;
+}
+
+function traversal($root, array &$output) {
+    if ($root->val === null) {
+        return;
+    }
+
+    $this->traversal($root->left, $output);
+    $output[] = $root->val;
+    $this->traversal($root->right, $output);
+}
+```
+```php
+// 145.后序遍历
+function postorderTraversal($root) {
+    $output = [];
+    $this->traversal($root, $output);
+    return $output;
+}
+
+function traversal($root, array &$output) {
+    if ($root->val === null) {
+        return;
+    }
+
+    $this->traversal($root->left, $output);
+    $this->traversal($root->right, $output);
+    $output[] = $root->val;
+}
+```
+
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">