Add solution #951

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,036 LeetCode solutions in JavaScript
+# 1,037 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -760,6 +760,7 @@
 948|[Bag of Tokens](./solutions/0948-bag-of-tokens.js)|Medium|
 949|[Largest Time for Given Digits](./solutions/0949-largest-time-for-given-digits.js)|Medium|
 950|[Reveal Cards In Increasing Order](./solutions/0950-reveal-cards-in-increasing-order.js)|Medium|
+951|[Flip Equivalent Binary Trees](./solutions/0951-flip-equivalent-binary-trees.js)|Medium|
 966|[Vowel Spellchecker](./solutions/0966-vowel-spellchecker.js)|Medium|
 970|[Powerful Integers](./solutions/0970-powerful-integers.js)|Easy|
 976|[Largest Perimeter Triangle](./solutions/0976-largest-perimeter-triangle.js)|Easy|

solutions/0951-flip-equivalent-binary-trees.js

@@ -0,0 +1,38 @@
+/**
+ * 951. Flip Equivalent Binary Trees
+ * https://leetcode.com/problems/flip-equivalent-binary-trees/
+ * Difficulty: Medium
+ *
+ * For a binary tree T, we can define a flip operation as follows: choose any node, and swap the
+ * left and right child subtrees.
+ *
+ * A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y
+ * after some number of flip operations.
+ *
+ * Given the roots of two binary trees root1 and root2, return true if the two trees are flip
+ * equivalent or false otherwise.
+ */
+
+/**
+ * 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} root1
+ * @param {TreeNode} root2
+ * @return {boolean}
+ */
+var flipEquiv = function(root1, root2) {
+  return helper(root1, root2);
+
+  function helper(node1, node2) {
+    if (!node1 && !node2) return true;
+    if (!node1 || !node2 || node1.val !== node2.val) return false;
+    return (helper(node1.left, node2.left) && helper(node1.right, node2.right))
+      || (helper(node1.left, node2.right) && helper(node1.right, node2.left));
+  }
+};