B-Tree 设计与实现总结--《算法导论》

总结自《introduction to algorithm》第3版，第18章的B-Tree。实现是用的java代码

• 定义
• B-Tree get 查
• B-tree insert 增
• 删除
• 完整代码

B-Tree其实是一种多路平衡树，主要是用在对辅存中的数据做增删改查，所以更大的时间消耗其实是在读写辅存。不过为了实现的简便我在实现中并不考虑辅存的读写。

定义

节点 x$x$ 的属性

1. x.n$x.n$ 关键字的个数
2. 每个关键字缝隙(x.keyi−x.keyi+1$x.ke{y}_i-x.ke{y}_{i+1}$) 之间有一个孩子节点，即共有x.n+1$x.n+1$ 个孩子（除叶子节点外）
3. x.isleaf 标识是否是叶子
4. 关键字有严格的递增顺序

此外，除了叶子节点外，每个节点的关键字都有一个上界和下界，我们称t$t$ 为B-Tree 的度，则，t−1≤x.n≤2t−1$t-1\le x.n\le 2t-1$
接下来我们来证明B-tree 的高度 h≤logtn+12$h\le lo{g}_t\frac{n+1}{2}$

节点的实现，我这里是将每个节点底层用两个数组来实现，一个entry，另外一个是childs，且大小固定为 2t−1，2t$2t-1，2t$

class TreeNode<K ,V>{
        Item<K,V>[] elememts;//关键字
        TreeNode<K,V>[] childs;//孩子节点
        int size;
        boolean isLeaf;
        @SuppressWarnings("unchecked")
        public TreeNode(){
            elememts = new Item[(degree <<1) -1];
            childs = new TreeNode[degree <<1];
            size =0;
            isLeaf = true;
        }
        public boolean isLeaf() {
            return isLeaf;
        }
        public final Item<K, V> last() {
            return elememts[size-1];
        }
        public final Item<K, V> first() {
            return elememts[0];
        }
        public final TreeNode<K, V> lastChild() {
            return childs[size];
        }
        public final TreeNode<K, V> firstChild() {
            return childs[0];
        }
    }

B-Tree 的高度

B-Tree 的根至少有t−q$t-q$ 个关键字，且至少每个节点的度数至少是 t$t$ 因此

n≥1+(t−1)∑i=1h2∗ti−1=2th−1h≤logtn+12$$\begin{gathered} n\ge 1+(t-1)\sum _{i=1}^{h}2\ast t^{i-1}=2t^h-1 \\ h\le lo{g}_t\frac{n+1}{2} \end{gathered}$$

B-Tree get 查

这是B-Tree的查询实现

public V get(K key) {
        TreeNode<K, V> cl = this.root;
        while (!cl.isLeaf) {//非叶子节点
            int idx = lowerBound(key, cl);// 找到严格大于key的item，返回值与 Arrays.binSearch类似
            if(idx>=0)return cl.elememts[idx].getValue();
            else cl = cl.childs[idx];//递归访问孩子
        }
        int idx = lowerBound(key, cl);
        if(idx<0)return null;
        else return cl.elememts[idx].value; 
    }

复杂度就是lowerBound查询每个节点的复杂度和树的深度 O(log t logtn)$O(logtlo{g}_{t}n)$

B-tree insert 增

插入一个关键字

插入关键字的时候，不能像二叉树搜索树一样直接创建一个新的叶子节点插入(这样会破坏B-Tree的性质)，而是插入到一个已经创建好的叶子节点上，这样也会出现一个问题:

当节点是满的的时候x.n=2t−1$x.n=2t-1$ 插入会破坏B-Tree的性质，因此我们引入一个新的操作，split, 将当前节点分裂成两个子节点，每个子节点各自 t−1$t-1$ 个，中间一个元素提升到父亲，当人正阳可能会引起新的分裂。我们递归的处理就好了，不过由于从根往下搜索的时候就能确定哪些节点是会分裂的，因此我们下搜的时候就将其分裂开来。

插入主代码

public boolean put(K key, V value) {
        if(root == null){
            root = new TreeNode<>();
            addItem(new Item<>(key, value), root, 0);
            return true;
        }
        TreeNode<K, V> r = this.root;
        if(isFull(r)){//r.size == 2*deg -1, 分裂树高+1
            TreeNode<K, V> newRoot = new TreeNode<>();
            this.root = newRoot;    
            addChild(r, newRoot, 0);
            splitChild(newRoot,0);
            return putNonFull(newRoot,new Item<>(key, value));
        }else return putNonFull(r, new Item<>(key, value)); 
    }
private boolean putNonFull(TreeNode<K, V> node, Item<K, V> entry) {
        int idx = lowerBound(entry, node);
        if(idx>=0){//修改value,return false
            node.elememts[idx].value = entry.value;
            return false;
        }
        idx = -idx -1;
        if(node.isLeaf){//叶子节点直接插入
            addItem(entry, node, idx);
        }else{
            TreeNode<K, V> lc = node.childs[idx];
            if(isFull(lc)){//分裂
                splitChild(node, idx);
                if(keyCmp.compare(entry, node.elememts[idx]) > 0)lc = node.childs[idx+1];
                putNonFull(lc, entry);
            }else putNonFull(lc, entry);
        }
        return true;
    }
private void splitChild(TreeNode<K, V> par, int idx){
        par.isLeaf =false;
        TreeNode<K, V> rc = new TreeNode<>();
        TreeNode<K, V> lc = par.childs[idx];
        assert isFull(lc); // lc.size == 2*degree - 1
        int sz =0;
        int deg = this.degree;
        while (sz < deg-1) {
            rc.elememts[sz] = lc.elememts[sz+deg];
            sz++;
        }
        rc.size = sz;
        if(!lc.isLeaf){
            rc.isLeaf = lc.isLeaf;
            while (sz>=0) {
                rc.childs[sz] = lc.childs[sz+deg];
                --sz;
            }
        }
        lc.size = deg -1;
        addItem(lc.elememts[deg-1], par, idx);
        addChild(rc, par, idx+1);
    }

删除

与插入对应，删除的时候也要引入一个操作，就是connect 链接，因为删除的时候我们保证每个节点至少有 t$t$ 个儿子(而不是B-tree 的t-1) 个。

算法导论中讨论了删除的时候的几种情况，但是没有提供伪代码，我这里按照它的思想实现了

可将代码对照着上面的图式看看

/*
     * 删除对应的节点，应用算法导论中的算法
     * key: 删除节点对应的key
     * return : false: 如果节点不存在，反之删除并且return true;
     */
    public boolean remove(K key) {
        TreeNode<K, V> cl = this.root;
        if(cl.size==0){//高度减一,根被删除，树高降低的唯一方式
            this.root = cl.childs[0];
            cl = this.root;
        }
        TreeNode<K, V> par = null;
        int degLowerBound = this.degree;
        while (!cl.isLeaf) {
            checkInvirantDelete(cl);
            int keyIdx = lowerBound(key, cl);
            if(keyIdx>=0){//key 在节点cl中
                TreeNode<K, V>lc = cl.childs[keyIdx],rc = cl.childs[keyIdx+1];
                if(lc.size >=degLowerBound){// case 2a递归删除 key的前驱，将前驱存在cl中，
                    Item<K, V> pre =  lc.last();
                    cl.elememts[keyIdx] = pre;
                    key = pre.key;
                    cl = lc;//递归删除左子树上的前驱节点;
                }else if (rc.size>=degLowerBound) {
                    Item<K, V> sus = rc.first();
                    cl.elememts[keyIdx] = sus;
                    key = sus.key;
                    cl = rc;
                }else{// 两颗子树都只有 deg-1个,合并 lc,keyItem,rc;
                    cl = connect(cl, keyIdx);               
                }
            }else{//case 3, 不在cl中
                keyIdx = - keyIdx -1;//lower bound
                TreeNode<K, V> now = cl.childs[keyIdx];
                if(now.size < degLowerBound){//仅有deg-1个关键字
                    TreeNode<K, V> brother;
                    if(keyIdx-1>=0 && (brother = cl.childs[keyIdx-1]).size>=degLowerBound){
                        TreeNode<K, V> adCl = brother.lastChild();// 移到now
                        addItem(cl.elememts[keyIdx], now, 0);
                        addChild(adCl, now, 0);
                        //将brother[last]移到 cl
                        cl.elememts[keyIdx] = deleteItemAndSon(brother, brother.size-1);
                    }else if(keyIdx+1 <=cl.size &&(brother = cl.childs[keyIdx+1]).size>=degLowerBound){
                        TreeNode<K, V> adCl = brother.firstChild();
                        addItem(cl.elememts[keyIdx], now, degLowerBound-1);
                        addChild(adCl, par, degLowerBound);
                        cl.elememts[keyIdx] = deleteItemAndSon(brother, 0);
                    }else {
                        //链接右兄弟
                        now = connect(cl, keyIdx);
                    }
                }
                cl = now;
            }
        }
        //cl为叶子节点
        int keyIdx = lowerBound(key, cl);
        if(keyIdx<0)return false;
        //删除叶子节点上的 key
        for(int bound = cl.size-1;keyIdx<bound; ++keyIdx)cl.elememts[keyIdx] = cl.elememts[keyIdx+1];
        --cl.size;
        return true;
    }

/*
     * 链接node.childs[idx] + node.ele[idx]+node.childs[idx+1]
     * node : 父亲节点
     * idx: 链接的元素位置
     * return: node.elements[idx]
     */
    private TreeNode<K, V> connect(TreeNode<K, V> node,int idx) {
        TreeNode<K, V> lc = node.childs[idx],rc = node.childs[idx+1];
        assert lc.size == this.degree-1 && rc.size == this.degree-1;
        int deg = this.degree;
        Item<K, V> ret = deleteItemAndSon(node, idx);
        lc.elememts[deg-1] = ret;
        for(int i=0 ;i<deg-1 ; ++i){
            lc.elememts[i+deg] = rc.elememts[i];
            lc.childs[i+deg] = rc.childs[i];
        }
        lc.childs[deg*2 -1] = rc.childs[deg-1];
        lc.size = 2*deg-1;
        return lc;
    }
/*
     * 删除node[idx] 对应位置的Item和son
     * node: 待删除节点
     * idx: 下标
     */
    private Item<K, V> deleteItemAndSon(TreeNode<K, V> node,int idx) {
        Item<K, V> ret = node.elememts[idx]; 
        int n = node.size;
        for(int i=idx ; i < n-1 ; ++i){
            node.elememts[i] = node.elememts[i+1];
            node.childs[i+1] = node.childs[i+2];
        }
        --node.size;
        return ret;
    }

完整代码

package dataStructure;

/*
 * 根据算法导论对B-Tree 的简单实现，主要在于应用算法，而不是考试实际的使用，因此为了简单默认key是可比较的，并且没有纳入java collection Framework
 * 主要实现以下算法:
 * delete
 * search
 * insert
 * get
 * connect
 * 即增删改查
 * author : zouzhitao
 */
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Map;
import java.util.Queue;

public class BTree<K extends Comparable<K>,V> {// 简化代码实现，默认 key是可以比较的
    private static final int DEFAULT_DEGREE = 3;// 默认b-tree的度数
    private final int degree;// b-tree 的度

    private TreeNode<K, V> root;
    private final Comparator<Item<K, V>> keyCmp = new Comparator<BTree.Item<K,V>>() {
        @Override
        public int compare(Item<K, V> o1, Item<K, V> o2) {
            return o1.key.compareTo(o2.key);
        }
    };


    public BTree(int degree) {
        super();
        this.degree = degree;
    }
    public BTree() {
        this.degree = DEFAULT_DEGREE;
    }

    public boolean put(K key, V value) {
        if(root == null){
            root = new TreeNode<>();
            addItem(new Item<>(key, value), root, 0);
            return true;
        }
        TreeNode<K, V> r = this.root;
        if(isFull(r)){//r.size == 2*deg -1, 分裂树高+1
            TreeNode<K, V> newRoot = new TreeNode<>();
            this.root = newRoot;    
            addChild(r, newRoot, 0);
            splitChild(newRoot,0);
            return putNonFull(newRoot,new Item<>(key, value));
        }else return putNonFull(r, new Item<>(key, value)); 
    }
    /*
     * 插入时候保证每个节点的size至少
     */
    private final void checkInvirantDelete(TreeNode<K, V> node) {
        boolean ok = node == this.root ||(node.size >=degree);
        assert ok: "不满足度数大于等于this.degree";
    }
    /*
     * 删除对应的节点，应用算法导论中的算法
     * key: 删除节点对应的key
     * return : false: 如果节点不存在，反之删除并且return true;
     */
    public boolean remove(K key) {
        TreeNode<K, V> cl = this.root;
        if(cl.size==0){//高度减一,根被删除，树高降低的唯一方式
            this.root = cl.childs[0];
            cl = this.root;
        }
        TreeNode<K, V> par = null;
        int degLowerBound = this.degree;
        while (!cl.isLeaf) {
            checkInvirantDelete(cl);
            int keyIdx = lowerBound(key, cl);
            if(keyIdx>=0){//key 在节点cl中
                TreeNode<K, V>lc = cl.childs[keyIdx],rc = cl.childs[keyIdx+1];
                if(lc.size >=degLowerBound){// case 2a递归删除 key的后继
                    Item<K, V> pre =  lc.last();
                    cl.elememts[keyIdx] = pre;
                    key = pre.key;
                    cl = lc;//递归删除左子树上的前驱节点;
                }else if (rc.size>=degLowerBound) {
                    Item<K, V> sus = rc.first();
                    cl.elememts[keyIdx] = sus;
                    key = sus.key;
                    cl = rc;
                }else{// 两颗子树都只有 deg-1个,合并 lc,keyItem,rc;
                    cl = connect(cl, keyIdx);               
                }
            }else{//case 3, 不在cl中
                keyIdx = - keyIdx -1;//lower bound
                TreeNode<K, V> now = cl.childs[keyIdx];
                if(now.size < degLowerBound){//仅有deg-1个关键字
                    TreeNode<K, V> brother;
                    if(keyIdx-1>=0 && (brother = cl.childs[keyIdx-1]).size>=degLowerBound){
                        TreeNode<K, V> adCl = brother.lastChild();// 移到now
                        addItem(cl.elememts[keyIdx], now, 0);
                        addChild(adCl, now, 0);
                        //将brother[last]移到 cl
                        cl.elememts[keyIdx] = deleteItemAndSon(brother, brother.size-1);
                    }else if(keyIdx+1 <=cl.size &&(brother = cl.childs[keyIdx+1]).size>=degLowerBound){
                        TreeNode<K, V> adCl = brother.firstChild();
                        addItem(cl.elememts[keyIdx], now, degLowerBound-1);
                        addChild(adCl, par, degLowerBound);
                        cl.elememts[keyIdx] = deleteItemAndSon(brother, 0);
                    }else {
                        //链接右兄弟
                        now = connect(cl, keyIdx);
                    }
                }
                cl = now;
            }
        }
        //cl为叶子节点
        int keyIdx = lowerBound(key, cl);
        if(keyIdx<0)return false;
        //删除叶子节点上的 key
        for(int bound = cl.size-1;keyIdx<bound; ++keyIdx)cl.elememts[keyIdx] = cl.elememts[keyIdx+1];
        --cl.size;
        return true;
    }
    /*
     * 链接node.childs[idx] + node.ele[idx]+node.childs[idx+1]
     * node : 父亲节点
     * idx: 链接的元素位置
     * return: node.elements[idx]
     */
    private TreeNode<K, V> connect(TreeNode<K, V> node,int idx) {
        TreeNode<K, V> lc = node.childs[idx],rc = node.childs[idx+1];
        assert lc.size == this.degree-1 && rc.size == this.degree-1;
        int deg = this.degree;
        Item<K, V> ret = deleteItemAndSon(node, idx);
        lc.elememts[deg-1] = ret;
        for(int i=0 ;i<deg-1 ; ++i){
            lc.elememts[i+deg] = rc.elememts[i];
            lc.childs[i+deg] = rc.childs[i];
        }
        lc.childs[deg*2 -1] = rc.childs[deg-1];
        lc.size = 2*deg-1;
        return lc;
    }
    /*
     * 删除node[idx] 对应位置的Item和son
     * node: 待删除节点
     * idx: 下标
     */
    private Item<K, V> deleteItemAndSon(TreeNode<K, V> node,int idx) {
        Item<K, V> ret = node.elememts[idx]; 
        int n = node.size;
        for(int i=idx ; i < n-1 ; ++i){
            node.elememts[i] = node.elememts[i+1];
            node.childs[i+1] = node.childs[i+2];
        }
        --node.size;
        return ret;
    }
    private boolean putNonFull(TreeNode<K, V> node, Item<K, V> entry) {
        int idx = lowerBound(entry, node);
        if(idx>=0){//修改value,return false
            node.elememts[idx].value = entry.value;
            return false;
        }
        idx = -idx -1;
        if(node.isLeaf){//叶子节点直接插入
            addItem(entry, node, idx);
        }else{
            TreeNode<K, V> lc = node.childs[idx];
            if(isFull(lc)){//分裂
                splitChild(node, idx);
                if(keyCmp.compare(entry, node.elememts[idx]) > 0)lc = node.childs[idx+1];
                putNonFull(lc, entry);
            }else putNonFull(lc, entry);
        }
        return true;
    }
    /*
     * 搜索key对应的value
     * return : return key对应的value 或者null,没有搜索到
     */
    public V get(K key) {
        TreeNode<K, V> cl = this.root;
        while (!cl.isLeaf) {
            int idx = lowerBound(key, cl);
            if(idx>=0)return cl.elememts[idx].getValue();
            else cl = cl.childs[idx];
        }
        int idx = lowerBound(key, cl);
        if(idx<0)return null;
        else return cl.elememts[idx].value; 
    }
    public String toString() {
        Queue<TreeNode<K, V>> Q=  new ArrayDeque<>();
        StringBuilder ret = new StringBuilder("root ");
        Q.add(this.root);
        int nodecnt = 0;
        while (!Q.isEmpty()) {
            TreeNode<K, V> vis = Q.remove();
            ret.append((nodecnt++)+" [ ");
            for(int i=0 ; i< vis.size ; ++i){
                Item<K, V> entry = vis.elememts[i];
                ret.append("( "+entry.key+ " = " + entry.value+" ),");
            }
            ret.append("]\n");
            if(!vis.isLeaf)for(int i=0 ; i<= vis.size ;++i)Q.add(vis.childs[i]);
        }
        return ret.toString();
    }
    private void splitChild(TreeNode<K, V> par, int idx){
        par.isLeaf =false;
        TreeNode<K, V> rc = new TreeNode<>();
        TreeNode<K, V> lc = par.childs[idx];
        assert isFull(lc); // lc.size == 2*degree - 1
        int sz =0;
        int deg = this.degree;
        while (sz < deg-1) {
            rc.elememts[sz] = lc.elememts[sz+deg];
            sz++;
        }
        rc.size = sz;
        if(!lc.isLeaf){
            rc.isLeaf = lc.isLeaf;
            while (sz>=0) {
                rc.childs[sz] = lc.childs[sz+deg];
                --sz;
            }
        }
        lc.size = deg -1;
        addItem(lc.elememts[deg-1], par, idx);
        addChild(rc, par, idx+1);
    }


    private void addItem(Item<K,V> entry,TreeNode<K, V> now, int idx) {
        for(int i = now.size-1 ; i >=idx ; --i)
            now.elememts[i+1] = now.elememts[i];
        now.elememts[idx] = entry;
        now.size++;
    }

    /*
     * 二分搜索第一个大于key的位置，如果key相等，返回-1;
     */
    private final int lowerBound(Item<K, V> key, TreeNode<K, V> now) {
        return Arrays.binarySearch(now.elememts,0,now.size,key, keyCmp);
    }
    private int lowerBound(K key, TreeNode<K, V> node) {
        Item<K, V> tmp = new Item<>(key);
        return lowerBound(tmp, node);
    }
    private void addChild(TreeNode<K, V> child, TreeNode<K, V> par, int idx) {
        assert idx <= par.size && par.size <= 2*degree-1;
        for(int i= par.size ; i >idx ; --i)
            par.childs[i] = par.childs[i-1];
        par.childs[idx] = child;
    }
    private boolean isFull(final TreeNode<K,V> now) {
        return now.size == this.degree * 2 -1;
    }

    static class Item<K,V> implements Map.Entry<K, V>{
        K key;
        V value;

        public Item(K key, V value) {
            super();
            this.key = key;
            this.value = value;
        }
        public Item(K key) {
            this.key = key;
        }
        @Override
        public K getKey() {
            return key;
        }

        @Override
        public V getValue() {
            return value;
        }

        @Override
        public V setValue(V value) {
            V oldValue = value;
            this.value = value;
            return oldValue;
        }
        public boolean equals(Object o) {
            if(!(o instanceof Item<?, ?>))
                return false;
            Item<?, ?> tmp = (Item<?, ?>)o;
            return valueEqual(this.key, tmp.key) && valueEqual(this.value,tmp.value);
        }
    }
    class TreeNode<K ,V>{
        Item<K,V>[] elememts;
        TreeNode<K,V>[] childs;
        int size;
        boolean isLeaf;
        @SuppressWarnings("unchecked")
        public TreeNode(){
            elememts = new Item[(degree <<1) -1];
            childs = new TreeNode[degree <<1];
            size =0;
            isLeaf = true;
        }
        public boolean isLeaf() {
            return isLeaf;
        }
        public final Item<K, V> last() {
            return elememts[size-1];
        }
        public final Item<K, V> first() {
            return elememts[0];
        }
        public final TreeNode<K, V> lastChild() {
            return childs[size];
        }
        public final TreeNode<K, V> firstChild() {
            return childs[0];
        }
    }
    private static final boolean valueEqual(Object o1,Object o2) {
        return o1==null? o2==null : o1.equals(o2);
    }

    public static void main(String[] args) {
        BTree<Character, Integer> map = new BTree<Character, Integer>();
        String keys = "ACDEGJKMNOPRSTUVXYZ";
        for(int i=0 ; i< keys.length() ; ++i){
            map.put(keys.charAt(i), i);
        }
        System.out.println("init btree\n\n"+map);
        map.put(keys.charAt(0), 3);
        System.out.println("after call put('A',3)\n\n"+map);
        map.remove('A');
        System.out.println("delete A\n\n"+map);     
        map.remove('B');
        System.out.println("affter call remove('B')\n"+map);

    }
}