From a45046be8a6387a0c4e7fbf6872c554b6a2f3249 Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Tue, 9 May 2023 15:11:13 -0500
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?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

添加 创建 root节点 和 python queue的代码优化版本
---
 problems/0617.合并二叉树.md | 67 ++++++++++++++++++++++++++++++--
 1 file changed, 64 insertions(+), 3 deletions(-)

diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md
index f98948f0..18a245c4 100644
--- a/problems/0617.合并二叉树.md
+++ b/problems/0617.合并二叉树.md
@@ -352,8 +352,7 @@ class Solution {
 ```
 
 ### Python 
-
-**递归法 - 前序遍历**
+(版本一) 递归 - 前序 - 修改root1
 ```python
 # Definition for a binary tree node.
 # class TreeNode:
@@ -377,8 +376,33 @@ class Solution:
         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. 
 
 ```
+(版本二) 递归 - 前序 - 新建root
+```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
+class Solution:
+    def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
+        # 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. 
+        root = TreeNode() # 创建新节点
+        root.val += root1.val + root2.val# 中
+        root.left = self.mergeTrees(root1.left, root2.left) #左
+        root.right = self.mergeTrees(root1.right, root2.right) # 右
+        
+        return root # ⚠️ 注意: 本题我们创建了新节点. 
 
-**迭代法**
+```
+
+(版本三) 迭代
 ```python
 class Solution:
     def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
@@ -413,7 +437,44 @@ class Solution:
 
         return root1
 ```
+(版本四) 迭代 + 代码优化
+```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
+from collections import deque
 
+class Solution:
+    def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
+        if not root1:
+            return root2
+        if not root2:
+            return root1
+
+        queue = deque()
+        queue.append((root1, root2))
+
+        while queue:
+            node1, node2 = queue.popleft()
+            node1.val += node2.val
+
+            if node1.left and node2.left:
+                queue.append((node1.left, node2.left))
+            elif not node1.left:
+                node1.left = node2.left
+
+            if node1.right and node2.right:
+                queue.append((node1.right, node2.right))
+            elif not node1.right:
+                node1.right = node2.right
+
+        return root1
+
+
+```
 ### Go 
 
 ```go