Add solution #1028

Author: JoshCrozier

Diff for: README.md

@@ -437,6 +437,7 @@
 1010|[Pairs of Songs With Total Durations Divisible by 60](./1010-pairs-of-songs-with-total-durations-divisible-by-60.js)|Medium|
 1022|[Sum of Root To Leaf Binary Numbers](./1022-sum-of-root-to-leaf-binary-numbers.js)|Easy|
 1023|[Camelcase Matching](./1023-camelcase-matching.js)|Medium|
+1028|[Recover a Tree From Preorder Traversal](./1028-recover-a-tree-from-preorder-traversal.js)|Hard|
 1037|[Valid Boomerang](./1037-valid-boomerang.js)|Easy|
 1041|[Robot Bounded In Circle](./1041-robot-bounded-in-circle.js)|Medium|
 1047|[Remove All Adjacent Duplicates In String](./1047-remove-all-adjacent-duplicates-in-string.js)|Easy|

Diff for: solutions/1028-recover-a-tree-from-preorder-traversal.js

@@ -0,0 +1,55 @@
+/**
+ * 1028. Recover a Tree From Preorder Traversal
+ * https://leetcode.com/problems/recover-a-tree-from-preorder-traversal/
+ * Difficulty: Hard
+ *
+ * We run a preorder depth-first search (DFS) on the root of a binary tree.
+ *
+ * At each node in this traversal, we output D dashes (where D is the depth of this node),
+ * then we output the value of this node.  If the depth of a node is D, the depth of its
+ * immediate child is D + 1.  The depth of the root node is 0.
+ *
+ * If a node has only one child, that child is guaranteed to be the left child.
+ *
+ * Given the output traversal of this traversal, recover the tree and return its root.
+ */
+
+/**
+ * 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 {string} traversal
+ * @return {TreeNode}
+ */
+var recoverFromPreorder = function(traversal) {
+  const stack = [];
+
+  for (let i = 0; i < traversal.length;) {
+    let depth = 0;
+    while (traversal[i] === '-') {
+      depth++;
+      i++;
+    }
+
+    let value = 0;
+    while (i < traversal.length && traversal[i] !== '-') {
+      value = value * 10 + parseInt(traversal[i]);
+      i++;
+    }
+
+    const node = new TreeNode(value);
+    while (stack.length > depth) stack.pop();
+    if (stack.length) {
+      if (!stack[stack.length - 1].left) stack[stack.length - 1].left = node;
+      else stack[stack.length - 1].right = node;
+    }
+    stack.push(node);
+  }
+
+  return stack[0];
+};