Add solution #623

Author: JoshCrozier

README.md

@@ -472,6 +472,7 @@
 617|[Merge Two Binary Trees](./0617-merge-two-binary-trees.js)|Easy|
 621|[Task Scheduler](./0621-task-scheduler.js)|Medium|
 622|[Design Circular Queue](./0622-design-circular-queue.js)|Medium|
+623|[Add One Row to Tree](./0623-add-one-row-to-tree.js)|Medium|
 628|[Maximum Product of Three Numbers](./0628-maximum-product-of-three-numbers.js)|Easy|
 637|[Average of Levels in Binary Tree](./0637-average-of-levels-in-binary-tree.js)|Easy|
 643|[Maximum Average Subarray I](./0643-maximum-average-subarray-i.js)|Easy|

solutions/0623-add-one-row-to-tree.js

@@ -0,0 +1,53 @@
+/**
+ * 623. Add One Row to Tree
+ * https://leetcode.com/problems/add-one-row-to-tree/
+ * Difficulty: Medium
+ *
+ * Given the root of a binary tree and two integers val and depth, add a row of nodes with value
+ * val at the given depth depth.
+ *
+ * Note that the root node is at depth 1.
+ *
+ * The adding rule is:
+ * - Given the integer depth, for each not null tree node cur at the depth depth - 1, create two
+ *   tree nodes with value val as cur's left subtree root and right subtree root.
+ * - cur's original left subtree should be the left subtree of the new left subtree root.
+ * - cur's original right subtree should be the right subtree of the new right subtree root.
+ * - If depth == 1 that means there is no depth depth - 1 at all, then create a tree node with
+ *   value val as the new root of the whole original tree, and the original tree is the new root's
+ *   left subtree.
+ */
+
+/**
+ * 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} root
+ * @param {number} val
+ * @param {number} depth
+ * @return {TreeNode}
+ */
+var addOneRow = function(root, val, depth) {
+  if (depth === 1) {
+    return new TreeNode(val, root);
+  }
+  dfs(root, 1);
+
+  return root;
+
+  function dfs(node, level) {
+    if (!node) return;
+    if (level === depth - 1) {
+      node.left = new TreeNode(val, node.left);
+      node.right = new TreeNode(val, null, node.right);
+      return;
+    }
+    dfs(node.left, level + 1);
+    dfs(node.right, level + 1);
+  }
+};