Add solution #897

Author: JoshCrozier

README.md

@@ -707,6 +707,7 @@
 894|[All Possible Full Binary Trees](./0894-all-possible-full-binary-trees.js)|Medium|
 895|[Maximum Frequency Stack](./0895-maximum-frequency-stack.js)|Hard|
 896|[Monotonic Array](./0896-monotonic-array.js)|Easy|
+897|[Increasing Order Search Tree](./0897-increasing-order-search-tree.js)|Easy|
 901|[Online Stock Span](./0901-online-stock-span.js)|Medium|
 905|[Sort Array By Parity](./0905-sort-array-by-parity.js)|Easy|
 909|[Snakes and Ladders](./0909-snakes-and-ladders.js)|Medium|

solutions/0897-increasing-order-search-tree.js

@@ -0,0 +1,38 @@
+/**
+ * 897. Increasing Order Search Tree
+ * https://leetcode.com/problems/increasing-order-search-tree/
+ * Difficulty: Easy
+ *
+ * Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost
+ * node in the tree is now the root of the tree, and every node has no left child and only one
+ * right child.
+ */
+
+/**
+ * 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 increasingBST = function(root) {
+  const result = new TreeNode(0);
+  let current = result;
+
+  inorderTraversal(root);
+  return result.right;
+
+  function inorderTraversal(node) {
+    if (!node) return;
+
+    inorderTraversal(node.left);
+    current.right = new TreeNode(node.val);
+    current = current.right;
+    inorderTraversal(node.right);
+  }
+};