Add solution #1382

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,266 LeetCode solutions in JavaScript
+# 1,267 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1053,6 +1053,7 @@
 1379|[Find a Corresponding Node of a Binary Tree in a Clone of That Tree](./solutions/1379-find-a-corresponding-node-of-a-binary-tree-in-a-clone-of-that-tree.js)|Easy|
 1380|[Lucky Numbers in a Matrix](./solutions/1380-lucky-numbers-in-a-matrix.js)|Easy|
 1381|[Design a Stack With Increment Operation](./solutions/1381-design-a-stack-with-increment-operation.js)|Medium|
+1382|[Balance a Binary Search Tree](./solutions/1382-balance-a-binary-search-tree.js)|Medium|
 1389|[Create Target Array in the Given Order](./solutions/1389-create-target-array-in-the-given-order.js)|Easy|
 1400|[Construct K Palindrome Strings](./solutions/1400-construct-k-palindrome-strings.js)|Medium|
 1402|[Reducing Dishes](./solutions/1402-reducing-dishes.js)|Hard|

solutions/1382-balance-a-binary-search-tree.js

@@ -0,0 +1,45 @@
+/**
+ * 1382. Balance a Binary Search Tree
+ * https://leetcode.com/problems/balance-a-binary-search-tree/
+ * Difficulty: Medium
+ *
+ * Given the root of a binary search tree, return a balanced binary search tree with the same
+ * node values. If there is more than one answer, return any of them.
+ *
+ * A binary search tree is balanced if the depth of the two subtrees of every node never differs
+ * by more than 1.
+ */
+
+/**
+ * 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 {TreeNode}
+ */
+var balanceBST = function(root) {
+  const nodes = [];
+  collectInOrder(root);
+  return buildBalancedTree(0, nodes.length - 1);
+
+  function collectInOrder(node) {
+    if (!node) return;
+    collectInOrder(node.left);
+    nodes.push(node.val);
+    collectInOrder(node.right);
+  }
+
+  function buildBalancedTree(start, end) {
+    if (start > end) return null;
+    const mid = Math.floor((start + end) / 2);
+    const node = new TreeNode(nodes[mid]);
+    node.left = buildBalancedTree(start, mid - 1);
+    node.right = buildBalancedTree(mid + 1, end);
+    return node;
+  }
+};