solve 1

Author: lylaminju

subtree-of-another-tree/lylaminju.py

@@ -0,0 +1,33 @@
+'''
+시간 복잡도: O(m * n)
+- m은 root 트리의 노드 수, n은 subRoot 트리의 노드 수
+
+공간 복잡도: O(h)
+- h는 root 트리의 높이
+'''
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+from typing import Optional
+
+
+class Solution:
+    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+        if not subRoot:
+            return True
+
+        if not root:
+            return False
+
+        return self.isSameTree(root, subRoot) or self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)
+    
+    def isSameTree(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
+        if not root1 and not root2:
+            return True
+
+        if not root1 or not root2:
+            return False
+
+        return root1.val == root2.val and self.isSameTree(root1.left, root2.left) and self.isSameTree(root1.right, root2.right)