Add solution #110

Author: JoshCrozier

README.md

@@ -103,6 +103,7 @@
 101|[Symmetric Tree](./0101-symmetric-tree.js)|Easy|
 102|[Binary Tree Level Order Traversal](./0102-binary-tree-level-order-traversal.js)|Medium|
 104|[Maximum Depth of Binary Tree](./0104-maximum-depth-of-binary-tree.js)|Easy|
+110|[Balanced Binary Tree](./0110-balanced-binary-tree.js)|Easy|
 111|[Minimum Depth of Binary Tree](./0111-minimum-depth-of-binary-tree.js)|Easy|
 112|[Path Sum](./0112-path-sum.js)|Easy|
 116|[Populating Next Right Pointers in Each Node](./0116-populating-next-right-pointers-in-each-node.js)|Medium|

solutions/0110-balanced-binary-tree.js

@@ -0,0 +1,30 @@
+/**
+ * 110. Balanced Binary Tree
+ * https://leetcode.com/problems/balanced-binary-tree/
+ * Difficulty: Easy
+ *
+ * Given a binary tree, determine if it is height-balanced.
+ */
+
+/**
+ * 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 {boolean}
+ */
+var isBalanced = function(root) {
+  if (!root) return true;
+  return isBalanced(root.right) && isBalanced(root.left)
+    && Math.abs(traverse(root.right) - traverse(root.left)) < 2;
+};
+
+function traverse(node, depth = 0) {
+  if (!node) return depth;
+  return Math.max(traverse(node.right, depth + 1), traverse(node.left, depth + 1));
+}