Add solution #173

Author: JoshCrozier

README.md

@@ -164,6 +164,7 @@
 168|[Excel Sheet Column Title](./0168-excel-sheet-column-title.js)|Easy|
 169|[Majority Element](./0169-majority-element.js)|Easy|
 171|[Excel Sheet Column Number](./0171-excel-sheet-column-number.js)|Easy|
+173|[Binary Search Tree Iterator](./0173-binary-search-tree-iterator.js)|Medium|
 179|[Largest Number](./0179-largest-number.js)|Medium|
 187|[Repeated DNA Sequences](./0187-repeated-dna-sequences.js)|Medium|
 189|[Rotate Array](./0189-rotate-array.js)|Medium|

solutions/0173-binary-search-tree-iterator.js

@@ -0,0 +1,58 @@
+/**
+ * 173. Binary Search Tree Iterator
+ * https://leetcode.com/problems/binary-search-tree-iterator/
+ * Difficulty: Medium
+ *
+ * Implement the BSTIterator class that represents an iterator over the in-order traversal
+ * of a binary search tree (BST):
+ * - BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of
+ *   the BST is given as part of the constructor. The pointer should be initialized to a
+ *   non-existent number smaller than any element in the BST.
+ * - boolean hasNext() Returns true if there exists a number in the traversal to the right
+ *   of the pointer, otherwise returns false.
+ * - int next() Moves the pointer to the right, then returns the number at the pointer.
+ *
+ * Notice that by initializing the pointer to a non-existent smallest number, the first call
+ * to next() will return the smallest element in the BST.
+ *
+ * You may assume that next() calls will always be valid. That is, there will be at least a
+ * next number in the in-order traversal when next() is called.
+ */
+
+/**
+ * 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
+ */
+var BSTIterator = function(root) {
+  this.stack = [];
+  this.root = root;
+};
+
+/**
+ * @return {number}
+ */
+BSTIterator.prototype.next = function() {
+  while (this.root) {
+    this.stack.push(this.root);
+    this.root = this.root.left;
+  }
+  this.root = this.stack.pop();
+
+  const result = this.root.val;
+  this.root = this.root.right;
+  return result;
+};
+
+/**
+ * @return {boolean}
+ */
+BSTIterator.prototype.hasNext = function() {
+  return this.root || this.stack.length;
+};