Add same-tree solution

same-tree/Jeehay28.js

@@ -0,0 +1,69 @@
+// 🚀 Iterative DFS (Stack)
+// ✅ Time Complexity: O(n), where n is the number of nodes in the tree
+// ✅ Space Complexity: O(n) (worst case), O(log n) (best case for balanced trees)
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {boolean}
+ */
+var isSameTree = function (p, q) {
+  let stack = [[p, q]];
+  while (stack.length > 0) {
+    const [p, q] = stack.pop();
+
+    if (p === null && q === null) continue;
+
+    if (p === null || q === null) return false;
+
+    if (p.val !== q.val) return false;
+
+    stack.push([p.left, q.left]);
+    stack.push([p.right, q.right]);
+  }
+
+  return true;
+};
+
+
+
+// 🚀 recursive approach
+// ✅ Time Complexity: O(n), where n is the number of nodes in the tree
+// ✅ Space Complexity: O(n) (worst case), O(log n) (best case for balanced trees)
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} p
+ * @param {TreeNode} q
+ * @return {boolean}
+ */
+// var isSameTree = function (p, q) {
+//   // Base case: If both trees are empty, they are the same
+//   if (p === null && q === null) return true;
+
+//   // If one of the trees is empty and the other is not, return false
+//   if (p === null || q === null) return false;
+
+//   // Compare the values of the current nodes
+//   if (p.val !== q.val) return false;
+
+//   // Recursively compare the left and right subtrees
+//   return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
+// };
+
+