Add solution #337

Author: JoshCrozier

README.md

@@ -245,6 +245,7 @@
 328|[Odd Even Linked List](./0328-odd-even-linked-list.js)|Medium|
 331|[Verify Preorder Serialization of a Binary Tree](./0331-verify-preorder-serialization-of-a-binary-tree.js)|Medium|
 334|[Increasing Triplet Subsequence](./0334-increasing-triplet-subsequence.js)|Medium|
+337|[House Robber III](./0337-house-robber-iii.js)|Medium|
 338|[Counting Bits](./0338-counting-bits.js)|Easy|
 342|[Power of Four](./0342-power-of-four.js)|Easy|
 344|[Reverse String](./0344-reverse-string.js)|Easy|

solutions/0337-house-robber-iii.js

@@ -0,0 +1,38 @@
+/**
+ * 337. House Robber III
+ * https://leetcode.com/problems/house-robber-iii/
+ * Difficulty: Medium
+ *
+ * The thief has found himself a new place for his thievery again. There is only one
+ * entrance to this area, called root.
+ *
+ * Besides the root, each house has one and only one parent house. After a tour, the smart
+ * thief realized that all houses in this place form a binary tree. It will automatically
+ * contact the police if two directly-linked houses were broken into on the same night.
+ *
+ * Given the root of the binary tree, return the maximum amount of money the thief can rob
+ * without alerting the police.
+ */
+
+/**
+ * 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
+ * @return {number}
+ */
+var rob = function(root) {
+  return Math.max(...traverse(root));
+
+  function traverse(node) {
+    if (!node) return [0, 0];
+    const [l1, l2] = traverse(node.left);
+    const [r1, r2] = traverse(node.right);
+    return [node.val + l2 + r2, Math.max(l1 + r1, l2 + r2, l1 + r2, l2 + r1)];
+  }
+};