Add solution #1372

Author: JoshCrozier

README.md

@@ -426,6 +426,7 @@
 1365|[How Many Numbers Are Smaller Than the Current Number](./1365-how-many-numbers-are-smaller-than-the-current-number.js)|Easy|
 1366|[Rank Teams by Votes](./1366-rank-teams-by-votes.js)|Medium|
 1368|[Minimum Cost to Make at Least One Valid Path in a Grid](./1368-minimum-cost-to-make-at-least-one-valid-path-in-a-grid.js)|Hard|
+1372|[Longest ZigZag Path in a Binary Tree](./1372-longest-zigzag-path-in-a-binary-tree.js)|Medium|
 1374|[Generate a String With Characters That Have Odd Counts](./1374-generate-a-string-with-characters-that-have-odd-counts.js)|Easy|
 1380|[Lucky Numbers in a Matrix](./1380-lucky-numbers-in-a-matrix.js)|Easy|
 1389|[Create Target Array in the Given Order](./1389-create-target-array-in-the-given-order.js)|Easy|

solutions/1372-longest-zigzag-path-in-a-binary-tree.js

@@ -0,0 +1,47 @@
+/**
+ * 1372. Longest ZigZag Path in a Binary Tree
+ * https://leetcode.com/problems/longest-zigzag-path-in-a-binary-tree/
+ * Difficulty: Medium
+ *
+ * You are given the root of a binary tree.
+ *
+ * A ZigZag path for a binary tree is defined as follow:
+ * - Choose any node in the binary tree and a direction (right or left).
+ * - If the current direction is right, move to the right child of the current node;
+ *   otherwise, move to the left child.
+ * - Change the direction from right to left or from left to right.
+ * - Repeat the second and third steps until you can't move in the tree.
+ *
+ * Zigzag length is defined as the number of nodes visited - 1. (A single node has a
+ * length of 0).
+ *
+ * Return the longest ZigZag path contained in that 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 {number}
+ */
+var longestZigZag = function(root) {
+  let result = 0;
+
+  dfs(root, true, 0);
+  dfs(root, false, 0);
+
+  return result;
+
+  function dfs(node, isLeft, total) {
+    if (!node) return;
+    result = Math.max(result, total);
+    dfs(node.left, true, isLeft ? 1 : total + 1);
+    dfs(node.right, false, isLeft ? total + 1 : 1);
+  }
+};