Merge pull request #534 from KelvinG-611/master

更改, 添加404.左叶子之和python部分代码

problems/0404.左叶子之和.md

@@ -205,25 +205,51 @@ class Solution {
 
 
 Python：
-```Python
+
 **递归**
-# 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
+```python
 class Solution:
     def sumOfLeftLeaves(self, root: TreeNode) -> int:
-        self.res=0
-        def areleftleaves(root):
-            if not root:return
-            if root.left and (not root.left.left) and (not root.left.right):self.res+=root.left.val
-            areleftleaves(root.left)
-            areleftleaves(root.right)
-        areleftleaves(root)
-        return self.res
+        if not root: 
+            return 0
+        
+        left_left_leaves_sum = self.sumOfLeftLeaves(root.left)  # 左
+        right_left_leaves_sum = self.sumOfLeftLeaves(root.right) # 右
+        
+        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
+            
+        return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum
 ```
+
+**迭代**
+```python
+class Solution:
+    def sumOfLeftLeaves(self, root: TreeNode) -> int:
+        """
+        Idea: Each time check current node's left node. 
+              If current node don't have one, skip it. 
+        """
+        stack = []
+        if root: 
+            stack.append(root)
+        res = 0
+        
+        while stack: 
+            # 每次都把当前节点的左节点加进去. 
+            cur_node = stack.pop()
+            if cur_node.left and not cur_node.left.left and not cur_node.left.right: 
+                res += cur_node.left.val
+                
+            if cur_node.left: 
+                stack.append(cur_node.left)
+            if cur_node.right: 
+                stack.append(cur_node.right)
+                
+        return res
+```
+
 Go：
 
 > 递归法