Add solution #99

Author: JoshCrozier

README.md

@@ -85,6 +85,7 @@
 95|[Unique Binary Search Trees II](./0095-unique-binary-search-trees-ii.js)|Medium|
 96|[Unique Binary Search Trees](./0096-unique-binary-search-trees.js)|Medium|
 98|[Validate Binary Search Tree](./0098-validate-binary-search-tree.js)|Medium|
+99|[Recover Binary Search Tree](./0099-recover-binary-search-tree.js)|Medium|
 100|[Same Tree](./0100-same-tree.js)|Easy|
 101|[Symmetric Tree](./0101-symmetric-tree.js)|Easy|
 102|[Binary Tree Level Order Traversal](./0102-binary-tree-level-order-traversal.js)|Medium|

solutions/0099-recover-binary-search-tree.js

@@ -0,0 +1,43 @@
+/**
+ * 99. Recover Binary Search Tree
+ * https://leetcode.com/problems/recover-binary-search-tree/
+ * Difficulty: Medium
+ *
+ * You are given the root of a binary search tree (BST), where the values
+ * of exactly two nodes of the tree were swapped by mistake. Recover the
+ * tree without changing its structure.
+ */
+
+/**
+ * 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 recoverTree = function(root) {
+  let [previous, small, large] = [null, null, null];
+
+  dfs(root);
+  [large.val, small.val] = [small.val, large.val];
+
+  function dfs(root) {
+    if (!root) return;
+    dfs(root.left);
+    if (previous != null && previous.val > root.val) {
+      small = root;
+      if (!large) {
+        large = previous;
+      } else {
+        return false;
+      }
+    }
+    previous = root;
+    dfs(root.right);
+  }
+};