Add solution #1008

Author: JoshCrozier

Diff for: README.md

@@ -1,4 +1,4 @@
-# 1,084 LeetCode solutions in JavaScript
+# 1,085 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -816,6 +816,7 @@
 1005|[Maximize Sum Of Array After K Negations](./solutions/1005-maximize-sum-of-array-after-k-negations.js)|Easy|
 1006|[Clumsy Factorial](./solutions/1006-clumsy-factorial.js)|Medium|
 1007|[Minimum Domino Rotations For Equal Row](./solutions/1007-minimum-domino-rotations-for-equal-row.js)|Medium|
+1008|[Construct Binary Search Tree from Preorder Traversal](./solutions/1008-construct-binary-search-tree-from-preorder-traversal.js)|Medium|
 1009|[Complement of Base 10 Integer](./solutions/1009-complement-of-base-10-integer.js)|Easy|
 1010|[Pairs of Songs With Total Durations Divisible by 60](./solutions/1010-pairs-of-songs-with-total-durations-divisible-by-60.js)|Medium|
 1022|[Sum of Root To Leaf Binary Numbers](./solutions/1022-sum-of-root-to-leaf-binary-numbers.js)|Easy|

Diff for: solutions/1008-construct-binary-search-tree-from-preorder-traversal.js

@@ -0,0 +1,45 @@
+/**
+ * 1008. Construct Binary Search Tree from Preorder Traversal
+ * https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/
+ * Difficulty: Medium
+ *
+ * Given an array of integers preorder, which represents the preorder traversal of a BST (i.e.,
+ * binary search tree), construct the tree and return its root.
+ *
+ * It is guaranteed that there is always possible to find a binary search tree with the given
+ * requirements for the given test cases.
+ *
+ * A binary search tree is a binary tree where for every node, any descendant of Node.left has
+ * a value strictly less than Node.val, and any descendant of Node.right has a value strictly
+ * greater than Node.val.
+ *
+ * A preorder traversal of a binary tree displays the value of the node first, then traverses
+ * Node.left, then traverses Node.right.
+ */
+
+/**
+ * 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 {number[]} preorder
+ * @return {TreeNode}
+ */
+var bstFromPreorder = function(preorder) {
+  let index = 0;
+  return buildBST(Infinity);
+
+  function buildBST(bound) {
+    if (index >= preorder.length || preorder[index] > bound) return null;
+
+    const node = new TreeNode(preorder[index++]);
+    node.left = buildBST(node.val);
+    node.right = buildBST(bound);
+
+    return node;
+  }
+};