【发布时间】：2021-01-13 10:05:13
【问题描述】：
在我学习理解 B+树的过程中，我现在想看看如何修改 this 现有的 B+ 索引树（它具有使每个数组的长度为 2 的幂的附加限制：1， 2, 4, 8, 16, 或 32), 并把它变成一个 Stack, 把它变成一个 Queue。这是来自精彩链接答案的原始 B+tree 代码：
class Node {
constructor(capacity) {
// Mimic fixed-size array (avoid accidentally growing it)
this.children = Object.seal(Array(capacity).fill(null));
this.childCount = 0; // Number of used slots in children array
this.treeSize = 0; // Total number of values in this subtree
// Maintain back-link to parent.
this.parent = null;
// Per level in the tree, maintain a doubly linked list
this.prev = this.next = null;
}
setCapacity(capacity) {
if (capacity < 1) return;
// Here we make a new array, and copy the data into it
let children = Object.seal(Array(capacity).fill(null));
for (let i = 0; i < this.childCount; i++) children[i] = this.children[i];
this.children = children;
}
isLeaf() {
return !(this.children[0] instanceof Node);
}
index() {
return this.parent.children.indexOf(this);
}
updateTreeSize(start, end, sign=1) {
let sum = 0;
if (this.isLeaf()) {
sum = end - start;
} else {
for (let i = start; i < end; i++) sum += this.children[i].treeSize;
}
if (!sum) return;
sum *= sign;
// Apply the sum change to this node and all its ancestors
for (let node = this; node; node = node.parent) {
node.treeSize += sum;
}
}
wipe(start, end) {
this.updateTreeSize(start, end, -1);
this.children.copyWithin(start, end, this.childCount);
for (let i = this.childCount - end + start; i < this.childCount; i++) {
this.children[i] = null;
}
this.childCount -= end - start;
// Reduce allocated size if possible
if (this.childCount * 2 <= this.children.length) this.setCapacity(this.children.length / 2);
}
moveFrom(neighbor, target, start, count=1) {
// Note: `start` can have two meanings:
// if neighbor is null, it is the value/Node to move to the target
// if neighbor is a Node, it is the index from where value(s) have to be moved to the target
// Make room in target node
if (this.childCount + count > this.children.length) this.setCapacity(this.children.length * 2);
this.children.copyWithin(target + count, target, Math.max(target + count, this.childCount));
this.childCount += count;
if (neighbor !== null) {
// Copy the children
for (let i = 0; i < count; i++) {
this.children[target + i] = neighbor.children[start + i];
}
// Remove the original references
neighbor.wipe(start, start + count);
} else {
this.children[target] = start; // start is value to insert
}
this.updateTreeSize(target, target + count, 1);
// Set parent link(s)
if (!this.isLeaf()) {
for (let i = 0; i < count; i++) {
this.children[target + i].parent = this;
}
}
}
moveToNext(count) {
this.next.moveFrom(this, 0, this.childCount - count, count);
}
moveFromNext(count) {
this.moveFrom(this.next, this.childCount, 0, count);
}
basicRemove(index) {
if (!this.isLeaf()) {
// Take node out of the level's linked list
let prev = this.children[index].prev;
let next = this.children[index].next;
if (prev) prev.next = next;
if (next) next.prev = prev;
}
this.wipe(index, index + 1);
}
basicInsert(index, value) {
this.moveFrom(null, index, value);
if (value instanceof Node) {
// Insert node in the level's linked list
if (index > 0) {
value.prev = this.children[index-1];
value.next = value.prev.next;
} else if (this.childCount > 1) {
value.next = this.children[1];
value.prev = value.next.prev;
}
if (value.prev) value.prev.next = value;
if (value.next) value.next.prev = value;
}
}
pairWithSmallest() {
return this.prev && (!this.next || this.next.childCount > this.prev.childCount)
? [this.prev, this] : [this, this.next];
}
toString() {
return "[" + this.children.map(v => v??"-").join() + "]";
}
}
class Tree {
constructor(nodeCapacity=32) {
this.nodeCapacity = nodeCapacity;
this.root = new Node(1);
this.first = this.root; // Head of doubly linked list at bottom level
}
locate(offset) {
let node = this.root;
// Normalise argument
offset = offset < 0 ? Math.max(0, node.treeSize + offset) : Math.min(offset, node.treeSize);
while (!node.isLeaf()) {
let index = 0;
let child = node.children[index];
while (offset > child.treeSize || offset === child.treeSize && child.next) {
offset -= child.treeSize;
child = node.children[++index];
}
node = child;
}
return [node, offset];
}
getItemAt(offset) {
let [node, index] = this.locate(offset);
if (index < node.childCount) return node.children[index];
}
setItemAt(offset, value) {
let [node, index] = this.locate(offset);
if (index < node.childCount) node.children[index] = value;
}
removeItemAt(offset) {
let [node, index] = this.locate(offset);
if (index >= node.childCount) return;
while (true) {
console.assert(node.isLeaf() || node.children[index].treeSize === 0);
node.basicRemove(index);
// Exit when node's fill ratio is fine
if (!node.parent || node.childCount * 2 > this.nodeCapacity) return;
// Node has potentially too few children, we should either merge or redistribute
let [left, right] = node.pairWithSmallest();
if (!left || !right) { // A node with no siblings? Must become the root!
this.root = node;
node.parent = null;
return;
}
let sumCount = left.childCount + right.childCount;
let childCount = sumCount >> 1;
// Check whether to merge or to redistribute
if (sumCount > this.nodeCapacity) { // redistribute
// Move some data from the bigger to the smaller node
let shift = childCount - node.childCount;
if (!shift) { // Boundary case: when a redistribution would bring no improvement
console.assert(node.childCount * 2 === this.nodeCapacity && sumCount === this.nodeCapacity + 1);
return;
}
if (node === left) { // move some children from right to left
left.moveFromNext(shift);
} else { // move some children from left to right
left.moveToNext(shift);
}
return;
}
// Merge:
// Move all data from the right to the left
left.moveFromNext(right.childCount);
// Prepare to delete right node
node = right.parent;
index = right.index();
}
}
insertItemAt(offset, value) {
let [node, index] = this.locate(offset);
while (node.childCount === this.nodeCapacity) { // No room here
if (index === 0 && node.prev && node.prev.childCount < this.nodeCapacity) {
return node.prev.basicInsert(node.prev.childCount, value);
}
// Check whether we can redistribute (to avoid a split)
if (node !== this.root) {
let [left, right] = node.pairWithSmallest();
let joinedIndex = left === node ? index : left.childCount + index;
let sumCount = left.childCount + right.childCount + 1;
if (sumCount <= 2 * this.nodeCapacity) { // redistribute
let childCount = sumCount >> 1;
if (node === right) { // redistribute to the left
let insertInLeft = joinedIndex < childCount;
left.moveFromNext(childCount - left.childCount - +insertInLeft);
} else { // redistribute to the right
let insertInRight = index >= sumCount - childCount;
left.moveToNext(childCount - right.childCount - +insertInRight);
}
if (joinedIndex > left.childCount ||
joinedIndex === left.childCount && left.childCount > right.childCount) {
right.basicInsert(joinedIndex - left.childCount, value);
} else {
left.basicInsert(joinedIndex, value);
}
return;
}
}
// Cannot redistribute: split node
let childCount = node.childCount >> 1;
// Create a new node that will later become the right sibling of this node
let sibling = new Node(childCount);
// Move half of node node's data to it
sibling.moveFrom(node, 0, childCount, childCount);
// Insert the value in either the current node or the new one
if (index > node.childCount) {
sibling.basicInsert(index - node.childCount, value);
} else {
node.basicInsert(index, value);
}
// Is this the root?
if (!node.parent) {
// ...then first create a parent, which is the new root
this.root = new Node(2);
this.root.basicInsert(0, node);
}
// Prepare for inserting the sibling node into the tree
index = node.index() + 1;
node = node.parent;
value = sibling;
}
node.basicInsert(index, value);
}
/* Below this point: these methods are optional */
* [Symbol.iterator]() { // Make tree iterable
let i = 0;
for (let node = this.first; node; node = node.next) {
for (let i = 0; i < node.childCount; i++) yield node.children[i];
}
}
print() {
console.log(this.root && this.root.toString());
}
verify() {
// Raise an error when the tree violates one of the required properties
if (!this.root) return; // An empty tree is fine.
if (this.root.parent) throw "root should not have a parent";
// Perform a breadth first traversal
let q = [this.root];
while (q.length) {
if (q[0].isLeaf() && this.first !== q[0]) throw "this.first is not pointing to first leaf";
let level = [];
let last = null;
for (let parent of q) {
if (!(parent instanceof Node)) throw "parent is not instance of Node";
if (parent.children.length > this.nodeCapacity) throw "node's children array is too large";
if (parent.childCount > 0 && parent.childCount * 2 <= parent.children.length) throw "node's fill ratio is too low";
for (let i = parent.childCount; i < parent.children.length; i++) {
if (parent.children[i] !== null) throw "child beyond childCount should be null but is not";
}
let treeSize = parent.treeSize;
if (parent.isLeaf()) {
for (let value of parent.children.slice(0, parent.childCount)) {
if (value === null) throw "leaf has a null as value";
if (value instanceof Node) throw "leaf has a Node as value";
}
if (parent.treeSize !== parent.childCount) throw "leaf has mismatch in treeSize and childCount";
} else {
for (let node of parent.children.slice(0, parent.childCount)) {
if (node === null) throw "internal node has a null as value";
if (!(node instanceof Node)) throw "internal node has a non-Node as value";
if (node.parent !== parent) throw "wrong parent";
if (node.prev !== last) throw "prev link incorrect";
if (last && last.next !== node) throw "next link incorrect";
if (last && last.children.length + node.children.length <= this.nodeCapacity) {
throw "two consecutive siblings have a total number of children that is too small";
}
if (node.childCount * 2 < this.nodeCapacity) {
throw "internal node is too small: " + node;
}
level.push(node);
last = node;
treeSize -= node.treeSize;
}
if (treeSize) throw "internal node treeSize sum mismatches";
}
}
if (last && last.next) throw "last node in level has a next reference";
q = level;
}
}
test(count=100, option=3) {
// option:
// 0 = always insert & delete at left side (offset 0)
// 1 = always insert & delete at right side
// 2 = always insert & delete at middle
// 3 = insert & delete at random offsets
// Create array to perform the same operations on it as on the tree
let arr = [];
// Perform a series of insertions
for (let i = 0; i < count; i++) {
// Choose random insertion index
let index = Array.isArray(option) ? option[i] : [0, i, i >> 1, Math.floor(Math.random() * (i+1))][option];
// Perform same insertion in array and tree
arr.splice(index, 0, i);
this.insertItemAt(index, i);
// Verify tree consistency and properties
this.verify();
// Verify the order of values in the array is the same as in the tree
if (arr+"" !== [...this]+"") throw i + ": tree not same as array";
}
// Perform a series of updates
for (let i = 0; i < count; i++) {
// Choose random update index
let index = Math.floor(Math.random() * count);
// Perform same insertion in array and tree
arr[index] += count;
this.setItemAt(index, this.getItemAt(index) + count);
// Verify tree consistency and properties
this.verify();
// Verify the order of values in the array is the same as in the tree
if (arr+"" !== [...this]+"") throw "tree not same as array";
}
// Perform a series of deletions
for (let i = arr.length - 1; i >= 0; i--) {
// Choose random deletion index
let index = [0, i, i >> 1, Math.floor(Math.random() * (i+1))][option];
// Perform same deletion in array and tree
arr.splice(index, 1);
this.removeItemAt(index);
// Verify tree consistency and properties
this.verify();
// Verify the order of values in the array is the same as in the tree
if (arr+"" !== [...this]+"") throw "tree not same as array";
}
}
}
// Perform 1000 insertions, 1000 updates, and 1000 deletions on a tree with node capacity of 8
new Tree(8).test(1000);
console.log("all tests completed");堆栈是后进先出的 LIFO 数据结构，而队列是先进先出的 FIFO 数据结构。堆栈具有push 和pop 方法（除了getItemAt(index) 和其他基本数组方法，此B+tree 已经实现）。队列有一个push 和shift 方法，其中shift 从数组的“前面”删除。所以它们已经很相似了，只需将项目添加到数组的末尾，或者从数组的开头或结尾删除。
我对现有 B+tree 执行 push 操作的方式是简单地跟踪数组的长度（您可以使用 tree.root.treeSize 进行操作），并在该位置进行 insertItemAt(tree.root.treeSize, val) 它。但也许有更优化的方式来做到这一点？
Tree.prototype.push = function(val) { this.insertItemAt(this.root.treeSize, val) }
对于流行音乐，我会这样做：
Tree.prototype.pop = function() {
let val = this.getItemAt(this.root.treeSize - 1)
this.removeItemAt(this.root.treeSize - 1)
return val
}
最后，对于shift，我会这样做：
Tree.prototype.shift = function() {
let val = this.getItemAt(0)
this.removeItemAt(0)
return val
}
但我的问题是，有没有更好更优化的方法来做到这一点？与其遍历整个树来查找第一个或最后一个项目，也许我们可以缓存它们？不确定这里的最佳方法。考虑到 B+tree 的结构（因为具有两个约束的强大功能，仅此而已），有什么方法可以使其达到最佳状态？例如，在“pluck”操作（pop 和 shift）中，有 两个遍历，也许这可能是一个（或者甚至没有）？如何修改此 B+树以使这些操作优化？
【问题讨论】：
标签： javascript arrays algorithm data-structures tree