Add solution #1361

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,254 LeetCode solutions in JavaScript
+# 1,255 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1035,6 +1035,7 @@
 1358|[Number of Substrings Containing All Three Characters](./solutions/1358-number-of-substrings-containing-all-three-characters.js)|Medium|
 1359|[Count All Valid Pickup and Delivery Options](./solutions/1359-count-all-valid-pickup-and-delivery-options.js)|Hard|
 1360|[Number of Days Between Two Dates](./solutions/1360-number-of-days-between-two-dates.js)|Easy|
+1361|[Validate Binary Tree Nodes](./solutions/1361-validate-binary-tree-nodes.js)|Medium|
 1365|[How Many Numbers Are Smaller Than the Current Number](./solutions/1365-how-many-numbers-are-smaller-than-the-current-number.js)|Easy|
 1366|[Rank Teams by Votes](./solutions/1366-rank-teams-by-votes.js)|Medium|
 1368|[Minimum Cost to Make at Least One Valid Path in a Grid](./solutions/1368-minimum-cost-to-make-at-least-one-valid-path-in-a-grid.js)|Hard|

solutions/1361-validate-binary-tree-nodes.js

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+/**
+ * 1361. Validate Binary Tree Nodes
+ * https://leetcode.com/problems/validate-binary-tree-nodes/
+ * Difficulty: Medium
+ *
+ * You have n binary tree nodes numbered from 0 to n - 1 where node i has two children
+ * leftChild[i] and rightChild[i], return true if and only if all the given nodes form
+ * exactly one valid binary tree.
+ *
+ * If node i has no left child then leftChild[i] will equal -1, similarly for the right child.
+ *
+ * Note that the nodes have no values and that we only use the node numbers in this problem.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[]} leftChild
+ * @param {number[]} rightChild
+ * @return {boolean}
+ */
+var validateBinaryTreeNodes = function(n, leftChild, rightChild) {
+  const parentCount = new Array(n).fill(0);
+  let rootCandidate = -1;
+
+  for (let i = 0; i < n; i++) {
+    if (leftChild[i] !== -1) parentCount[leftChild[i]]++;
+    if (rightChild[i] !== -1) parentCount[rightChild[i]]++;
+  }
+
+  for (let i = 0; i < n; i++) {
+    if (parentCount[i] === 0) {
+      if (rootCandidate !== -1) return false;
+      rootCandidate = i;
+    }
+    if (parentCount[i] > 1) return false;
+  }
+
+  if (rootCandidate === -1) return false;
+
+  return countNodes(rootCandidate, new Set()) === n;
+
+  function countNodes(node, visited) {
+    if (node === -1) return 0;
+    if (visited.has(node)) return -1;
+    visited.add(node);
+    const leftNodes = countNodes(leftChild[node], visited);
+    const rightNodes = countNodes(rightChild[node], visited);
+    if (leftNodes === -1 || rightNodes === -1) return -1;
+    return 1 + leftNodes + rightNodes;
+  }
+};