From 84a12f32de997868efcfff08d25c6a161fcc5cff Mon Sep 17 00:00:00 2001
From: Kelvin <guanli611@outlook.com>
Date: Sat, 31 Jul 2021 23:10:30 -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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迭代法: 逻辑没有改, 只是改变顺序使之与c++代码逻辑一致.
---
 problems/0617.合并二叉树.md | 51 ++++++++++++++++++--------------
 1 file changed, 29 insertions(+), 22 deletions(-)

diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md
index c5eb7594..09d844f3 100644
--- a/problems/0617.合并二叉树.md
+++ b/problems/0617.合并二叉树.md
@@ -340,31 +340,38 @@ class Solution:
 
 **迭代法**
 ```python
-# 迭代法-覆盖原来的树      
 class Solution:
     def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
-        if not root1: return root2
-        if not root2: return root1
-        # 迭代,将树2覆盖到树1
-        queue1 = [root1]
-        queue2 = [root2]
-        root = root1
-        while queue1 and queue2:
-            root1 = queue1.pop(0)
-            root2 = queue2.pop(0)
-            root1.val += root2.val
-            if not root1.left: # 如果树1左儿子不存在，则覆盖后树1的左儿子为树2的左儿子
-                root1.left = root2.left
-            elif root1.left and root2.left:
-                queue1.append(root1.left)
-                queue2.append(root2.left)
+        if not root1: 
+            return root2
+        if not root2: 
+            return root1
 
-            if not root1.right: # 同理，处理右儿子
-                root1.right = root2.right
-            elif root1.right and root2.right:
-                queue1.append(root1.right)
-                queue2.append(root2.right)
-        return root
+        queue = deque()
+        queue.append(root1)
+        queue.append(root2)
+
+        while queue: 
+            node1 = queue.popleft()
+            node2 = queue.popleft()
+            # 更新queue
+            # 只有两个节点都有左节点时, 再往queue里面放.
+            if node1.left and node2.left: 
+                queue.append(node1.left)
+                queue.append(node2.left)
+            # 只有两个节点都有右节点时, 再往queue里面放.
+            if node1.right and node2.right: 
+                queue.append(node1.right)
+                queue.append(node2.right)
+
+            # 更新当前节点. 同时改变当前节点的左右孩子. 
+            node1.val += node2.val
+            if not node1.left and node2.left: 
+                node1.left = node2.left
+            if not node1.right and node2.right: 
+                node1.right = node2.right
+
+        return root1
 ```
 
 Go：