feat: solve #228 with python

Author: EgonD3V

same-tree/EGON.py

@@ -0,0 +1,57 @@
+from typing import Optional
+from unittest import TestCase, main
+
+
+# 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 isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        return self.solve_dfs(p, q)
+
+    """
+    Runtime: 0 ms (Beats 100.00%)
+    Time Complexity: O(min(p, q))
+    > dfs를 통해 모든 node를 방문하므로, 각 트리의 node의 갯수를 각각 p, q라 하면, O(min(p, q))
+
+    Memory: 16.62 MB (Beats 15.78%)
+    Space Complexity: O(min(p, q))
+    > 일반적인 경우 트리의 깊이만큼 dfs 호출 스택이 쌓이나, 최악의 경우 한쪽으로 편향되었다면 O(min(p, q)), upper bound
+    """
+    def solve_dfs(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+
+        def dfs(p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+            if p is None and q is None:
+                return True
+            elif (p is not None and q is not None) and (p.val == q.val):
+                return dfs(p.left, q.left) and dfs(p.right, q.right)
+            else:
+                return False
+
+        return dfs(p, q)
+
+
+class _LeetCodeTestCases(TestCase):
+    def test_1(self):
+        p_1 = TreeNode(1)
+        p_2 = TreeNode(2)
+        p_3 = TreeNode(3)
+        p_1.left = p_2
+        p_1.right = p_3
+
+        q_1 = TreeNode(1)
+        q_2 = TreeNode(3)
+        q_3 = TreeNode(3)
+        q_1.left = q_2
+        q_1.right = q_3
+
+        self.assertEqual(Solution().isSameTree(p_1, q_1), True)
+
+
+if __name__ == '__main__':
+    main()