https://leetcode.cn/problems/closest-nodes-queries-in-a-binary-search-tree

Author: masx200

README.md

@@ -45,6 +45,8 @@ Step 2. Add the dependency
 
 <summary>展开查看</summary>
 
+https://leetcode.cn/problems/closest-nodes-queries-in-a-binary-search-tree
+
 https://leetcode.cn/problems/number-of-unequal-triplets-in-array
 
 https://leetcode.cn/problems/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance/

closest-nodes-queries-in-a-binary-search-tree/index.ts

@@ -0,0 +1,19 @@
+import { TreeNode } from "../binary-tree-inorder-traversal/TreeNode.ts";
+import inorderTraversal from "../binary-tree-inorder-traversal/index.ts";
+import { lowerBound } from "../maximum-profit-in-job-scheduling/lowerBound.ts";
+import { upperBound } from "../maximum-profit-in-job-scheduling/upperBound.ts";
+
+export default function closestNodes(
+    root: TreeNode | null,
+    queries: number[],
+): number[][] {
+    const nums = inorderTraversal(root);
+
+    return queries.map((v) => {
+        const left = lowerBound(0, nums.length, (m) => nums[m] <= v);
+        const mini = nums[left - 1] <= v ? nums[left - 1] : -1;
+        const right = upperBound(0, nums.length, (m) => nums[m] >= v);
+        const maxi = nums[right] >= v ? nums[right] : -1;
+        return [mini, maxi];
+    });
+}

number-of-unequal-triplets-in-array/index.ts

@@ -1,15 +1,15 @@
-export default function unequalTriplets(nums: number[]): number {
-    nums.sort((a, b) => a - b);
-
-    let ans = 0;
-    let start = 0;
-    const n = nums.length;
-    for (let i = 0; i < n - 1; i++) {
-        const x = nums[i];
-        if (x != nums[i + 1]) {
-            ans += start * (i - start + 1) * (n - 1 - i);
-            start = i + 1;
-        }
-    }
-    return ans;
-}
+export default function unequalTriplets(nums: number[]): number {
+    nums.sort((a, b) => a - b);
+
+    let ans = 0;
+    let start = 0;
+    const n = nums.length;
+    for (let i = 0; i < n - 1; i++) {
+        const x = nums[i];
+        if (x != nums[i + 1]) {
+            ans += start * (i - start + 1) * (n - 1 - i);
+            start = i + 1;
+        }
+    }
+    return ans;
+}