diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md
index e55981e2..19149285 100644
--- a/problems/0404.左叶子之和.md
+++ b/problems/0404.左叶子之和.md
@@ -42,7 +42,7 @@ if (node->left != NULL && node->left->left == NULL && node->left->right == NULL)
 
 ## 递归法
 
-递归的遍历顺序为后序遍历（左右中），是因为要通过递归函数的返回值来累加求取左叶子数值之和。。
+递归的遍历顺序为后序遍历（左右中），是因为要通过递归函数的返回值来累加求取左叶子数值之和。（前序遍历其实也同样AC）
 
 递归三部曲：
 
@@ -230,8 +230,8 @@ class Solution {
 
 ## Python 
 
-**递归**
-```python
+> 递归后序遍历
+```python3
 class Solution:
     def sumOfLeftLeaves(self, root: TreeNode) -> int:
         if not root: 
@@ -246,9 +246,36 @@ class Solution:
             
         return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum
 ```
+> 递归前序遍历
+```python3
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def sumOfLeftLeaves(self, root: TreeNode) -> int:
+        # 需要通过中节点来判断其的左节点是否存在；左节点自己的左右节点也是否存在
 
-**迭代**
-```python
+        if not root: return 0
+
+        # 初始化left_leaf备用
+        left_leaf = 0
+        # 若当前节点的左孩子就是左叶子
+        if root.left and not root.left.left and not root.left.right:
+            left_leaf = root.left.val
+
+        left_left_leaves_sum = self.sumOfLeftLeaves(root.left)
+        right_left_leaves_sum = self.sumOfLeftLeaves(root.right)
+
+
+        return left_leaf + left_left_leaves_sum + right_left_leaves_sum
+```
+
+
+> 迭代
+```python3
 class Solution:
     def sumOfLeftLeaves(self, root: TreeNode) -> int:
         """