diff --git a/617_merge_two_binary_trees/memo.md b/617_merge_two_binary_trees/memo.md
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+# 617. Merge Two Binary Trees
+
+https://leetcode.com/problems/merge-two-binary-trees/description/
+
+## Comments
+
+### step1
+
+*   最初 iterative DFS (`nodes_pairs = [(root1, root2)]` のような stack) で書こうとしたが、結局、今の node だけに注目していると、どの node の child なのか管理しないといけない気がして、再帰に方針変更
+    *   limit が 2000 なので、まあ recursionlimit を変更するかは環境次第か
+*   10:00 くらいで `Solution1` まで書いてみた。`new_node.left = merge_trees_helper(node1.left, node2.left)` のあたりで None に left / right をやって runtime error になった。この時点で、node1 が None、node2 が None、両方 None でないの 3 パターンがあり、これごとに条件分岐するの非常に面倒だなと思う
+*   get_left もしくは三項演算子を使って node1_left みたいなのを定義はできるが、node1_left, node2_left. node1_right, node2_right すべてやるの面倒だな…
+*   ここまでで 13:00 以上経過していたので一旦打ち切り
+*   -> あとで一応ベタ書きしてみた…これはひどい (`Solution2`)。二分木ではなく四分木だったらもっとひどい。
+
+### step2
+
+*   https://github.com/hayashi-ay/leetcode/pull/12/files
+    *   なるほど既存の木を編集していく想定だった？
+    *   ああそうか片方が None のときは merge する必要がなく、もう片方を返せばいいだけか。
+    *   そうすると merge するのは、いずれも None でないケースに限定されるので書きやすい
+    *   ここなんとなく書きたくはなるが確かに `if root1 is None` でカバーされているから不要ではある
+    *   `çopy.deepcopy` で非破壊的にできる。コピーに時間かかるがいいかは微妙だけど。
+
+```python
+        if root1 is None and root2 is None:
+            return None
+```
+
+*   https://github.com/nittoco/leetcode/pull/30/files#r1693985989
+    *   これでシンプルに書ける
+*   https://github.com/nittoco/leetcode/pull/30/files#r1693945861
+    *   仕事を押し付ける、押し付けないの議論がよく理解できていないのでまた読む。
+
+### step3
+
+*   2:00 前後。とりあえずこれが一番シンプルな解法だと思われる。
+*   ここは却って読みにくい気もするので、あまり使わないほうがいいかな。
+
+```python
+        # if root1 is None or root2 is None:
+        #     return root1 or root2
+```
diff --git a/617_merge_two_binary_trees/step1.py b/617_merge_two_binary_trees/step1.py
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+++ b/617_merge_two_binary_trees/step1.py
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+# 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
+# Runtime error
+class Solution1:
+    def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:
+        def merge_trees_helper(node1, node2):
+            if node1 is None and node2 is None:
+                return None
+            node1_val = node1.val if node1 is not None else 0
+            node2_val = node2.val if node2 is not None else 0
+            new_node = TreeNode(node1_val + node2_val)
+            new_node.left = merge_trees_helper(node1.left, node2.left)
+            new_node.right = merge_trees_helper(node1.right, node2.right)
+            return new_node
+
+        return merge_trees_helper(root1, root2)
+
+
+# ベタ書き。これはひどい
+# Accepted
+class Solution2:
+    def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:
+        def merge_trees_helper(node1, node2):
+            if node1 is None and node2 is None:
+                return None
+            node1_val = node1.val if node1 is not None else 0
+            node2_val = node2.val if node2 is not None else 0
+            new_node = TreeNode(node1_val + node2_val)
+            if node1 is not None and node2 is not None:
+                node1_left = node1.left
+                node1_right = node1.right
+                node2_left = node2.left
+                node2_right = node2.right
+            elif node1 is None:
+                node1_left = None
+                node1_right = None
+                node2_left = node2.left
+                node2_right = node2.right
+            elif node2 is None:
+                node1_left = node1.left
+                node1_right = node1.right
+                node2_left = None
+                node2_right = None
+        
+            new_node.left = merge_trees_helper(node1_left, node2_left)
+            new_node.right = merge_trees_helper(node1_right, node2_right)
+            return new_node
+
+        return merge_trees_helper(root1, root2)
diff --git a/617_merge_two_binary_trees/step2.py b/617_merge_two_binary_trees/step2.py
new file mode 100644
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+++ b/617_merge_two_binary_trees/step2.py
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+# 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: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:
+        if root1 is None and root2 is None:
+            return None
+        if root1 is None:
+            return root2
+        if root2 is None:
+            return root1
+        
+        new_root = TreeNode(root1.val + root2.val)
+        new_root.left = self.mergeTrees(root1.left, root2.left)
+        new_root.right = self.mergeTrees(root1.right, root2.right)
+        return new_root
diff --git a/617_merge_two_binary_trees/step3.py b/617_merge_two_binary_trees/step3.py
new file mode 100644
index 0000000..ff8ea4c
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+++ b/617_merge_two_binary_trees/step3.py
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+# 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: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:
+        # if root1 is None or root2 is None:
+        #     return root1 or root2
+        if root1 is None:
+            return root2
+        if root2 is None:
+            return root1
+        
+        new_root = TreeNode(root1.val + root2.val)
+        new_root.left = self.mergeTrees(root1.left, root2.left)
+        new_root.right = self.mergeTrees(root1.right, root2.right)
+        return new_root