Add solution #114

Author: JoshCrozier

Diff for: README.md

@@ -108,6 +108,7 @@
 111|[Minimum Depth of Binary Tree](./0111-minimum-depth-of-binary-tree.js)|Easy|
 112|[Path Sum](./0112-path-sum.js)|Easy|
 113|[Path Sum II](./0113-path-sum-ii.js)|Medium|
+114|[Flatten Binary Tree to Linked List](./0114-flatten-binary-tree-to-linked-list.js)|Medium|
 116|[Populating Next Right Pointers in Each Node](./0116-populating-next-right-pointers-in-each-node.js)|Medium|
 118|[Pascal's Triangle](./0118-pascals-triangle.js)|Easy|
 119|[Pascal's Triangle II](./0119-pascals-triangle-ii.js)|Easy|

Diff for: solutions/0114-flatten-binary-tree-to-linked-list.js

@@ -0,0 +1,39 @@
+/**
+ * 114. Flatten Binary Tree to Linked List
+ * https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
+ * Difficulty: Medium
+ *
+ * Given the root of a binary tree, flatten the tree into a "linked list":
+ * - The "linked list" should use the same TreeNode class where the right child pointer points
+ *   to the next node in the list and the left child pointer is always null.
+ * - The "linked list" should be in the same order as a pre-order traversal of the binary tree.
+ */
+
+/**
+ * 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 {void} Do not return anything, modify root in-place instead.
+ */
+var flatten = function(root) {
+  if (root === null) return;
+
+  if (root.left) {
+    let previous = root.left;
+    while (previous.right) {
+      previous = previous.right;
+    }
+    const rightNode = root.right;
+    root.right = root.left;
+    previous.right = rightNode;
+    root.left = null;
+  }
+
+  flatten(root.right);
+};