Fix borrowFromRight: replace silent warn with error log + add linked-list/separator invariant tests

utils/__tests__/bPlusTreeLogic.test.ts

@@ -390,6 +390,51 @@ describe('BPlusTree', () => {
     });
   });
 
+  describe('链表完整性 - borrowFromRight 场景', () => {
+    it('借用右兄弟后叶子 next 链表指针应保持完整', () => {
+      const tree = new BPlusTree((r: TableRow) => r.id, { uniqueKeys: true });
+      // Insert [1..7] then delete 1 to trigger borrowFromRight on the leftmost leaf
+      const rows = Array.from({ length: 7 }, (_, i) => ({ id: i + 1, name: `U${i + 1}`, age: 20 + i }));
+      rows.forEach(r => tree.insert(r));
+      tree.delete(1);
+
+      // Traverse linked list from leftmost leaf
+      let leaf: any = tree.root;
+      while (!leaf.isLeaf) leaf = leaf.children[0];
+
+      const keys: number[] = [];
+      while (leaf) {
+        keys.push(...leaf.keys);
+        leaf = leaf.next;
+      }
+
+      // All remaining keys should be reachable in sorted order
+      expect(keys).toEqual([2, 3, 4, 5, 6, 7]);
+    });
+
+    it('大量插入删除后 next 链路应完整且有序', () => {
+      const tree = new BPlusTree((r: TableRow) => r.id, { uniqueKeys: true });
+      const rows = Array.from({ length: 20 }, (_, i) => ({ id: i + 1, name: `U${i + 1}`, age: 20 + i }));
+      rows.forEach(r => tree.insert(r));
+
+      // Delete 10 keys spread across the tree
+      [2, 4, 6, 8, 10, 12, 14, 16, 18, 20].forEach(k => tree.delete(k));
+
+      // Traverse linked list from leftmost leaf
+      let leaf: any = tree.root;
+      while (!leaf.isLeaf) leaf = leaf.children[0];
+
+      const keys: number[] = [];
+      while (leaf) {
+        keys.push(...leaf.keys);
+        leaf = leaf.next;
+      }
+
+      // Only odd ids should remain, in sorted order
+      expect(keys).toEqual([1, 3, 5, 7, 9, 11, 13, 15, 17, 19]);
+    });
+  });
+
   describe('分隔键不变量', () => {
     it('删除叶子首key但不下溢时应更新父分隔键', () => {
       const tree = new BPlusTree((r: TableRow) => r.id, { uniqueKeys: true });
@@ -421,6 +466,16 @@ describe('BPlusTree', () => {
 
       assertSeparatorInvariants(tree.root);
     });
+
+    it('借用右兄弟后父分隔键应满足不变量', () => {
+      const tree = new BPlusTree((r: TableRow) => r.id, { uniqueKeys: true });
+      // Insert [1..7] then delete 1 to trigger borrowFromRight
+      const rows = Array.from({ length: 7 }, (_, i) => ({ id: i + 1, name: `U${i + 1}`, age: 20 + i }));
+      rows.forEach(r => tree.insert(r));
+      tree.delete(1);
+
+      assertSeparatorInvariants(tree.root);
+    });
   });
 });
 

utils/bPlusTreeLogic.ts

@@ -282,12 +282,19 @@ export class BPlusTree {
       const borrowedKey = rightSibling.keys.shift()!;
       const borrowedData = rightSibling.data.shift()!;
 
-      // Update parent's separator key to the new minimum key in right sibling
+      // After borrowing, update the separator key in parent to the new minimum key
+      // of the right sibling (i.e., its first remaining key).
+      // This maintains the B+ tree invariant: parent.keys[childIdx] == rightSibling.min
       if (rightSibling.keys.length > 0) {
         parent.keys[childIdx] = rightSibling.keys[0];
       } else {
-        // Right sibling became empty - should trigger merge instead
-        console.warn('Right sibling became empty after borrow, consider merging instead');
+        // This should never happen: handleUnderflow only calls borrowFromRight when
+        // rightSibling.keys.length > MIN_KEYS, guaranteeing at least 2 keys before borrow.
+        // If it does happen, the tree structure is already corrupted; log a clear error.
+        console.error(
+          '[BPlusTree] borrowFromRight: right sibling became empty after borrow. ' +
+          'This indicates a bug in handleUnderflow. Tree structure may be corrupted.'
+        );
       }
 
       node.keys.push(borrowedKey);