Update 0617.合并二叉树.md

修改逻辑与题解相同.

Author: KelvinG-LGTM

problems/0617.合并二叉树.md

@@ -312,23 +312,34 @@ class Solution {
 ```
 
 Python：
+
+**递归法 - 前序遍历**
 ```python
 # 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
-# 递归法*前序遍历
 class Solution:
     def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
-        if not root1: return root2  // 如果t1为空，合并之后就应该是t2
-        if not root2: return root1  // 如果t2为空，合并之后就应该是t1
-        root1.val = root1.val + root2.val  //中
-        root1.left = self.mergeTrees(root1.left , root2.left)  //左
-        root1.right = self.mergeTrees(root1.right , root2.right)  //右
-        return root1  //root1修改了结构和数值
+        # 递归终止条件: 
+        #  但凡有一个节点为空, 就立刻返回另外一个. 如果另外一个也为None就直接返回None. 
+        if not root1: 
+            return root2
+        if not root2: 
+            return root1
+        # 上面的递归终止条件保证了代码执行到这里root1, root2都非空. 
+        root1.val += root2.val # 中
+        root1.left = self.mergeTrees(root1.left, root2.left) #左
+        root1.right = self.mergeTrees(root1.right, root2.right) # 右
+        
+        return root1 # ⚠️ 注意: 本题我们重复使用了题目给出的节点而不是创建新节点. 节省时间, 空间. 
 
+```
+
+**迭代法**
+```python
 # 迭代法-覆盖原来的树      
 class Solution:
     def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode: