Merge pull request #545 from KelvinG-611/617.合并二叉树

Update 0617.合并二叉树.md

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:
@@ -319,41 +321,57 @@ 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
         
-# 迭代法-覆盖原来的树      
+        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
 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：