solve: subtreeOfAnotherTree

Author: yolophg

subtree-of-another-tree/yolophg.py

@@ -0,0 +1,36 @@
+# Time Complexity: O(m * n) m: number of nodes in root, n: number of nodes in subRoot, n: number of nodes in subRoot - might call the check function (which is O(n)) on every node in root.
+# Space Complexity: O(n) - use a queue for BFS that can hold up to O(n) nodes in the worst case.
+
+class Solution:
+    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+        if root is None:
+            return False 
+        if subRoot is None:
+            return True
+
+        # helper to compare two trees
+        def check(node1, node2):
+            if node1 is None and node2 is None:
+                return True 
+            if node1 is None or node2 is None:
+                return False 
+            if node1.val != node2.val:
+                return False 
+            # check left and right recursively
+            return check(node1.left, node2.left) and check(node1.right, node2.right)
+
+        # BFS through the main tree
+        queue = [root]
+        while queue:
+            curr = queue.pop(0)
+            # if value matches subRoot, check deeper
+            if curr.val == subRoot.val:
+                if check(curr, subRoot):
+                    return True
+            # add child nodes to keep exploring
+            if curr.left:
+                queue.append(curr.left)
+            if curr.right:
+                queue.append(curr.right)
+
+        return False