Add solution #426

Author: JoshCrozier

README.md

@@ -414,6 +414,7 @@
 423|[Reconstruct Original Digits from English](./solutions/0423-reconstruct-original-digits-from-english.js)|Medium|
 424|[Longest Repeating Character Replacement](./solutions/0424-longest-repeating-character-replacement.js)|Medium|
 425|[Word Squares](./solutions/0425-word-squares.js)|Hard|
+426|[Convert Binary Search Tree to Sorted Doubly Linked List](./solutions/0426-convert-binary-search-tree-to-sorted-doubly-linked-list.js)|Medium|
 427|[Construct Quad Tree](./solutions/0427-construct-quad-tree.js)|Medium|
 429|[N-ary Tree Level Order Traversal](./solutions/0429-n-ary-tree-level-order-traversal.js)|Medium|
 430|[Flatten a Multilevel Doubly Linked List](./solutions/0430-flatten-a-multilevel-doubly-linked-list.js)|Medium|

solutions/0426-convert-binary-search-tree-to-sorted-doubly-linked-list.js

@@ -0,0 +1,58 @@
+/**
+ * 426. Convert Binary Search Tree to Sorted Doubly Linked List
+ * https://leetcode.com/problems/convert-binary-search-tree-to-sorted-doubly-linked-list/
+ * Difficulty: Medium
+ *
+ * Convert a Binary Search Tree to a sorted Circular Doubly-Linked List in place.
+ *
+ * You can think of the left and right pointers as synonymous to the predecessor and successor
+ * pointers in a doubly-linked list. For a circular doubly linked list, the predecessor of the
+ * first element is the last element, and the successor of the last element is the first element.
+ *
+ * We want to do the transformation in place. After the transformation, the left pointer of the
+ * tree node should point to its predecessor, and the right pointer should point to its successor.
+ * You should return the pointer to the smallest element of the linked list.
+ */
+
+/**
+ * // Definition for a _Node.
+ * function _Node(val, left, right) {
+ *      this.val = val;
+ *      this.left = left;
+ *      this.right = right;
+ *  };
+ */
+
+/**
+ * @param {_Node} root
+ * @return {_Node}
+ */
+var treeToDoublyList = function(root) {
+  if (!root) return null;
+
+  let first = null;
+  let last = null;
+
+  inorder(root);
+
+  last.right = first;
+  first.left = last;
+
+  return first;
+
+  function inorder(node) {
+    if (!node) return;
+
+    inorder(node.left);
+
+    if (last) {
+      last.right = node;
+      node.left = last;
+    } else {
+      first = node;
+    }
+    last = node;
+
+    inorder(node.right);
+  }
+};