From 932d27754cb34420926a41176f6fedb860768b44 Mon Sep 17 00:00:00 2001
From: Kelvin <guanli611@outlook.com>
Date: Sat, 31 Jul 2021 23:04:46 -0400
Subject: [PATCH] =?UTF-8?q?Update=200617.=E5=90=88=E5=B9=B6=E4=BA=8C?=
 =?UTF-8?q?=E5=8F=89=E6=A0=91.md?=
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修改逻辑与题解相同.
---
 problems/0617.合并二叉树.md | 25 ++++++++++++++++++-------
 1 file changed, 18 insertions(+), 7 deletions(-)

diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md
index 19b58bd3..c5eb7594 100644
--- a/problems/0617.合并二叉树.md
+++ b/problems/0617.合并二叉树.md
@@ -312,6 +312,8 @@ class Solution {
 ```
 
 Python：
+
+**递归法 - 前序遍历**
 ```python
 # Definition for a binary tree node.
 # class TreeNode:
@@ -319,16 +321,25 @@ Python：
 #         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: