[선재] WEEK 15 Solutions

subtree-of-another-tree/sunjae95.js

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+/**
+ * @description
+ * root tree의 각 노드들에서 subRoot의 root와 같으면 preOrder를 통해 일치하는지 확인하는 로직
+ *
+ * n = count of root node
+ * time complexity: O(n^2)
+ * space complexity: O(n^2)
+ */
+var isSubtree = function (root, subRoot) {
+  const findTree = (tree, target) => {
+    if (!tree && !target) return true;
+    if (!tree || !target || tree.val !== target.val) return false;
+
+    if (!findTree(tree.left, target.left)) return false;
+    if (!findTree(tree.right, target.right)) return false;
+
+    return true;
+  };
+
+  const preOrder = (tree) => {
+    if (!tree) return false;
+
+    if (tree.val === subRoot.val) {
+      if (findTree(tree, subRoot)) return true;
+    }
+    if (preOrder(tree.left)) return true;
+    if (preOrder(tree.right)) return true;
+
+    return false;
+  };
+
+  return preOrder(root);
+};