feat: Solve subtree-of-another-tree problem

Author: hu6r1s

subtree-of-another-tree/hu6r1s.py

@@ -0,0 +1,24 @@
+# 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 isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+        if not subRoot:
+            return True
+        if not root:
+            return False
+
+        def same(root, subRoot):
+            if not root or not subRoot:
+                return not root and not subRoot
+
+            if root.val != subRoot.val:
+                return False
+            return same(root.left, subRoot.left) and same(root.right, subRoot.right)
+
+        if same(root, subRoot):
+            return True
+        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)