Update 0404.左叶子之和.md

勘误python注释

Author: casnz1601

problems/0404.左叶子之和.md

@@ -171,10 +171,10 @@ class Solution {
         int rightValue = sumOfLeftLeaves(root.right);  // 右
 
         int midValue = 0;
-        if (root.left != null && root.left.left == null && root.left.right == null) { // 中
+        if (root.left != null && root.left.left == null && root.left.right == null) { 
             midValue = root.left.val;
         }
-        int sum = midValue + leftValue + rightValue;
+        int sum = midValue + leftValue + rightValue;  // 中
         return sum;
     }
 }
@@ -242,37 +242,10 @@ class Solution:
 
         cur_left_leaf_val = 0
         if root.left and not root.left.left and not root.left.right: 
-            cur_left_leaf_val = root.left.val  # 中
+            cur_left_leaf_val = root.left.val 
 
-        return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum
+        return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum # 中
 ```
-> 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
-
-        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