Add solution #1245

Author: JoshCrozier

README.md

@@ -1155,6 +1155,7 @@
 1240|[Tiling a Rectangle with the Fewest Squares](./solutions/1240-tiling-a-rectangle-with-the-fewest-squares.js)|Hard|
 1243|[Array Transformation](./solutions/1243-array-transformation.js)|Easy|
 1244|[Design A Leaderboard](./solutions/1244-design-a-leaderboard.js)|Medium|
+1245|[Tree Diameter](./solutions/1245-tree-diameter.js)|Medium|
 1247|[Minimum Swaps to Make Strings Equal](./solutions/1247-minimum-swaps-to-make-strings-equal.js)|Medium|
 1248|[Count Number of Nice Subarrays](./solutions/1248-count-number-of-nice-subarrays.js)|Medium|
 1249|[Minimum Remove to Make Valid Parentheses](./solutions/1249-minimum-remove-to-make-valid-parentheses.js)|Medium|

solutions/1245-tree-diameter.js

@@ -0,0 +1,60 @@
+/**
+ * 1245. Tree Diameter
+ * https://leetcode.com/problems/tree-diameter/
+ * Difficulty: Medium
+ *
+ * The diameter of a tree is the number of edges in the longest path in that tree.
+ *
+ * There is an undirected tree of n nodes labeled from 0 to n - 1. You are given a 2D array
+ * edges where edges.length == n - 1 and edges[i] = [ai, bi] indicates that there is an
+ * undirected edge between nodes ai and bi in the tree.
+ *
+ * Return the diameter of the tree.
+ */
+
+/**
+ * @param {number[][]} edges
+ * @return {number}
+ */
+var treeDiameter = function(edges) {
+  if (edges.length === 0) return 0;
+
+  const graph = new Map();
+  for (const [a, b] of edges) {
+    if (!graph.has(a)) graph.set(a, []);
+    if (!graph.has(b)) graph.set(b, []);
+    graph.get(a).push(b);
+    graph.get(b).push(a);
+  }
+
+  const [farthestFromStart] = bfs(0);
+  const [, diameter] = bfs(farthestFromStart);
+
+  return diameter;
+
+  function bfs(start) {
+    const visited = new Set();
+    const queue = [[start, 0]];
+    visited.add(start);
+    let farthestNode = start;
+    let maxDistance = 0;
+
+    while (queue.length > 0) {
+      const [node, distance] = queue.shift();
+
+      if (distance > maxDistance) {
+        maxDistance = distance;
+        farthestNode = node;
+      }
+
+      for (const neighbor of graph.get(node) || []) {
+        if (!visited.has(neighbor)) {
+          visited.add(neighbor);
+          queue.push([neighbor, distance + 1]);
+        }
+      }
+    }
+
+    return [farthestNode, maxDistance];
+  }
+};