diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md
index dd9155c6..3a776837 100644
--- a/problems/0027.移除元素.md
+++ b/problems/0027.移除元素.md
@@ -106,6 +106,37 @@ public:
 
 旧文链接：[数组：就移除个元素很难么？](https://programmercarl.com/0027.移除元素.html)
 
+```CPP
+/**
+* 相向双指针方法，基于元素顺序可以改变的题目描述改变了元素相对位置，确保了移动最少元素
+* 时间复杂度：$O(n)$
+* 空间复杂度：$O(1)$
+*/
+class Solution {
+public:
+    int removeElement(vector<int>& nums, int val) {
+        int leftIndex = 0;
+        int rightIndex = nums.size() - 1;
+        while (leftIndex <= rightIndex) {
+            // 找左边等于val的元素
+            while (leftIndex <= rightIndex && nums[leftIndex] != val){
+                ++leftIndex;
+            }
+            // 找右边不等于val的元素
+            while (leftIndex <= rightIndex && nums[rightIndex] == val) {
+                -- rightIndex;
+            }
+            // 将右边不等于val的元素覆盖左边等于val的元素
+            if (leftIndex < rightIndex) {
+                nums[leftIndex++] = nums[rightIndex--];
+            }
+        }
+        return leftIndex;   // leftIndex一定指向了最终数组末尾的下一个元素
+    }
+};
+```
+
+  
 ## 相关题目推荐
 
 * 26.删除排序数组中的重复项
diff --git a/problems/0513.找树左下角的值.md b/problems/0513.找树左下角的值.md
index 84ed3932..12c62c70 100644
--- a/problems/0513.找树左下角的值.md
+++ b/problems/0513.找树左下角的值.md
@@ -433,6 +433,51 @@ var findBottomLeftValue = function(root) {
 };
 ```
 
+## TypeScript
+
+> 递归法：
+
+```typescript
+function findBottomLeftValue(root: TreeNode | null): number {
+    function recur(root: TreeNode, depth: number): void {
+        if (root.left === null && root.right === null) {
+            if (depth > maxDepth) {
+                maxDepth = depth;
+                resVal = root.val;
+            }
+            return;
+        }
+        if (root.left !== null) recur(root.left, depth + 1);
+        if (root.right !== null) recur(root.right, depth + 1);
+    }
+    let maxDepth: number = 0;
+    let resVal: number = 0;
+    if (root === null) return resVal;
+    recur(root, 1);
+    return resVal;
+};
+```
+
+> 迭代法：
+
+```typescript
+function findBottomLeftValue(root: TreeNode | null): number {
+    let helperQueue: TreeNode[] = [];
+    if (root !== null) helperQueue.push(root);
+    let resVal: number = 0;
+    let tempNode: TreeNode;
+    while (helperQueue.length > 0) {
+        resVal = helperQueue[0].val;
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            if (tempNode.left !== null) helperQueue.push(tempNode.left);
+            if (tempNode.right !== null) helperQueue.push(tempNode.right);
+        }
+    }
+    return resVal;
+};
+```
+
 ## Swift 
 
 递归版本：
diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md
index 468f2675..5e9fbdfe 100644
--- a/problems/0701.二叉搜索树中的插入操作.md
+++ b/problems/0701.二叉搜索树中的插入操作.md
@@ -310,6 +310,26 @@ class Solution:
         return root
 ``` 
 
+**递归法** - 无返回值 - another easier way
+```python
+class Solution:
+    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
+        newNode = TreeNode(val)
+        if not root: return newNode
+        
+        if not root.left and val < root.val:
+            root.left = newNode
+        if not root.right and val > root.val:
+            root.right = newNode
+        
+        if val < root.val:
+            self.insertIntoBST(root.left, val)
+        if val > root.val:
+            self.insertIntoBST(root.right, val)
+        
+        return root
+```
+
 **迭代法**  
 与无返回值的递归函数的思路大体一致 
 ```python