集合框架知识系列06 HashMap和TreeMap中的红黑树

在上一节中,HashMap在jdk 1.8中用了链表和红黑树两种方式解决冲突,在TreeMap中也是用红黑树存储的。下面分析一下红黑树的结构和基本操作。

一、红黑树的特征和基本操作

上一节中已经描述了红黑树的基本概念和特征,下面直接通过一个例子分析红黑树的构造和调整方法。

1、红黑树的数据结构

红黑树是一棵二叉查找树,在二叉树的基础上增加了节点的颜色,下面是TreeMap中的红黑树定义:
private static final boolean RED   = false;
private static final boolean BLACK = true;
static final class Entry<K,V> implements Map.Entry<K,V> {
        K key;
        V value;
        Entry<K,V> left;
        Entry<K,V> right;
        Entry<K,V> parent;
        boolean color = BLACK;

        /**
         * 给定key、value和父节点,构造一个新的。其中节点颜色为黑色
         */
        Entry(K key, V value, Entry<K,V> parent) {
            this.key = key;
            this.value = value;
            this.parent = parent;
        }
}

2、红黑树的左旋和右旋

红黑树的插入和删除,都有可能破坏其特性,就不是一棵红黑树了,所以要调整。调整的方法又两种,一种是改变某个节点的颜色,另外一种是结构调整,包括左旋和右旋。
左旋:将X的节点的右儿子节点Y变为其父节点,并且将Y的左子树变为X的右子树,变换过程入下图

《集合框架知识系列06 HashMap和TreeMap中的红黑树》

右旋:将X的节点的左儿子节点Y变为其父节点,并且将Y的右子树变为X的左子树,变换过程入下图

《集合框架知识系列06 HashMap和TreeMap中的红黑树》

3、插入节点后调整红黑树

当在红黑树中插入一个节点后,可能会破坏红黑树的规则,首先再回顾一下红黑数的特点:

  • 节点是红色或黑色。
  • 根节点是黑色。
  • 每个叶子节点都是黑色的空节点(NIL节点)。
  • 每个红色节点的两个子节点都是黑色。(从每个叶子到根的所有路径上不能有两个连续的红色节点)
  • 从任一节点到其每个叶子的所有路径都包含相同数目的黑色节点。

    从上面的条件可以看出,a肯定是不会违背的。插入的节点不在根节点处,所以b也不会违背。插入的节点时非空节点,c也不会违背。最有可能违背的就是d和e。而在我们插入节点时,先将要插入的节点颜色设置为红色,这样也就不会违背e。所以,插入后只需要调整不违背e就可以。
    插入后调整需要分三种情况来处理:

  1. 插入的是根节点:

    处理方法是直接将根节点颜色设置为黑色
    
  2. 插入节点的父节点为黑色节点或父节点为根节点

    不需要处理

  3. 插入节点的父节点时红色节点

    这种又分为三种情况
    下面假设插入节点为x,父节点为xp,祖父节点为xpp,祖父节点的左儿子为xppl,祖父节点的右儿子为xppr

  • S1:当前节点的父节点xp是红色,且当前节点的祖父节xpp点的另一个子节点(xppl或者xppr)也是红色

    处理逻辑:将父节点xp设为红色,祖父节点的儿子节点(xppl或者xppr)设为黑色,将祖父节点xpp设为红色,将祖父节点xpp设为当前节点,继续处理。

  • S2:当前节点的父节点xp是红色,祖父节点的儿子节点(xppl或者xppr)是黑色,且当前节点x是其父节点xp的右孩子

    处理逻辑:父节点xp作为当前节点x, 以当前节点x为支点进行左旋。

  • S3:当前节点的父节点xp是红色,祖父节点的儿子节点(xppl或者xppr)是黑色,且当前节点是其父节点xp的左孩子

    处理逻辑:将父节点xp设置为黑色,祖父节点xpp设置为红色,以祖父节点xpp为支点进行右旋

4、删除节点后调整红黑树

未完,待续。。。

3、构造一棵红黑树

  1. 通过插入节点,构造红黑树
    现在给定节点8 5 3 9 12 1 4 2,依次插入红黑树中,具体流程见下图:

《集合框架知识系列06 HashMap和TreeMap中的红黑树》

  1. 在红黑树中删除节点

未完,待续。。。

二、HashMap中的红黑树相关源码

static final class TreeNode<K,V> extends LinkedHashMap.Entry<K,V> {
        TreeNode<K,V> parent;  // red-black tree links
        TreeNode<K,V> left;
        TreeNode<K,V> right;
        //在节点删除后,需解除链接
        TreeNode<K,V> prev; 
        boolean red;
        TreeNode(int hash, K key, V val, Node<K,V> next) {
            super(hash, key, val, next);
        }

        /**
         * 返回根节点
         */
        final TreeNode<K,V> root() {
            for (TreeNode<K,V> r = this, p;;) {
                if ((p = r.parent) == null)
                    return r;
                r = p;
            }
        }

        /**
         * 确保根节点就是第一个节点
         */
        static <K,V> void moveRootToFront(Node<K,V>[] tab, TreeNode<K,V> root) {
            int n;
            if (root != null && tab != null && (n = tab.length) > 0) {
                int index = (n - 1) & root.hash;
                TreeNode<K,V> first = (TreeNode<K,V>)tab[index];
                //如果根节点不是第一个节点,进行调整
                if (root != first) {
                    Node<K,V> rn;
                    tab[index] = root;
                    TreeNode<K,V> rp = root.prev;
                    if ((rn = root.next) != null)
                        ((TreeNode<K,V>)rn).prev = rp;
                    if (rp != null)
                        rp.next = rn;
                    if (first != null)
                        first.prev = root;
                    root.next = first;
                    root.prev = null;
                }
                assert checkInvariants(root);
            }
        }

        /**
         * 根据hash值和key查询节点
         */
        final TreeNode<K,V> find(int h, Object k, Class<?> kc) {
            TreeNode<K,V> p = this;
            do {
                int ph, dir; K pk;
                TreeNode<K,V> pl = p.left, pr = p.right, q;
                if ((ph = p.hash) > h)
                    p = pl;
                else if (ph < h)
                    p = pr;
                else if ((pk = p.key) == k || (k != null && k.equals(pk)))
                    return p;
                else if (pl == null)
                    p = pr;
                else if (pr == null)
                    p = pl;
                else if ((kc != null ||
                          (kc = comparableClassFor(k)) != null) &&
                         (dir = compareComparables(kc, k, pk)) != 0)
                    p = (dir < 0) ? pl : pr;
                else if ((q = pr.find(h, k, kc)) != null)
                    return q;
                else
                    p = pl;
            } while (p != null);
            return null;
        }

        /**
         * 根据hash值和key查询节点
         */
        final TreeNode<K,V> getTreeNode(int h, Object k) {
            return ((parent != null) ? root() : this).find(h, k, null);
        }

        /**
         * 将链表转换为红黑树
         */
        final void treeify(Node<K,V>[] tab) {
            TreeNode<K,V> root = null;
               //从第一个节点开始
            for (TreeNode<K,V> x = this, next; x != null; x = next) {
                next = (TreeNode<K,V>)x.next;
                x.left = x.right = null;
                //如果root节点为null,x为根节点,此节点为黑色,父节点为null
                if (root == null) {
                    x.parent = null;
                    x.red = false;
                    root = x;
                }
                else {
                  //x的key值
                    K k = x.key;
                   //x的hash值
                    int h = x.hash;
                    Class<?> kc = null;
                    for (TreeNode<K,V> p = root;;) {
                        int dir, ph;
                        K pk = p.key;
                        //左边
                        if ((ph = p.hash) > h)
                            dir = -1;
                        //右边
                        else if (ph < h)
                            dir = 1;
                        //通过仲裁方法判断
                        else if ((kc == null &&
                                  (kc = comparableClassFor(k)) == null) ||
                                 (dir = compareComparables(kc, k, pk)) == 0)
                            dir = tieBreakOrder(k, pk);

                        TreeNode<K,V> xp = p;
                        //dir <=0 左子树搜索,并且判断左儿子是否为空,表示是否到叶子节点
                        if ((p = (dir <= 0) ? p.left : p.right) == null) {
                            x.parent = xp;
                            if (dir <= 0)
                                xp.left = x;
                            else
                                xp.right = x;
                            //插入元素,判断是否平衡,并且调整
                            root = balanceInsertion(root, x);
                            break;
                        }
                    }
                }
            }
          //确保根节点就是第一个节点
            moveRootToFront(tab, root);
        }

        /**
         * 红黑树转换为链表
         */
        final Node<K,V> untreeify(HashMap<K,V> map) {
            Node<K,V> hd = null, tl = null;
            for (Node<K,V> q = this; q != null; q = q.next) {
                Node<K,V> p = map.replacementNode(q, null);
                if (tl == null)
                    hd = p;
                else
                    tl.next = p;
                tl = p;
            }
            return hd;
        }

        /**
         * 插入一个节点
         */
        final TreeNode<K,V> putTreeVal(HashMap<K,V> map, Node<K,V>[] tab,
                                       int h, K k, V v) {
            Class<?> kc = null;
            boolean searched = false;
            TreeNode<K,V> root = (parent != null) ? root() : this;
            //从根据点开始,和当前搜索节点的hash比较
            for (TreeNode<K,V> p = root;;) {
                int dir, ph; K pk;
                if ((ph = p.hash) > h)
                    dir = -1;
                else if (ph < h)
                    dir = 1;
                 //hash和key都一致
                else if ((pk = p.key) == k || (k != null && k.equals(pk)))
                    return p;
                else if ((kc == null &&
                          (kc = comparableClassFor(k)) == null) ||
                         (dir = compareComparables(kc, k, pk)) == 0) {
                    if (!searched) {
                        TreeNode<K,V> q, ch;
                        searched = true;
                        if (((ch = p.left) != null &&
                             (q = ch.find(h, k, kc)) != null) ||
                            ((ch = p.right) != null &&
                             (q = ch.find(h, k, kc)) != null))
                            return q;
                    }
                    dir = tieBreakOrder(k, pk);
                }

                TreeNode<K,V> xp = p;
                if ((p = (dir <= 0) ? p.left : p.right) == null) {
                    Node<K,V> xpn = xp.next;
                    //新建节点
                    TreeNode<K,V> x = map.newTreeNode(h, k, v, xpn);
                    if (dir <= 0)
                        xp.left = x;
                    else
                        xp.right = x;
                    xp.next = x;
                    x.parent = x.prev = xp;
                    if (xpn != null)
                        ((TreeNode<K,V>)xpn).prev = x;
                   //插入元素,判断是否平衡,并且调整。确保根节点就是第一个节点
                    moveRootToFront(tab, balanceInsertion(root, x));
                    return null;
                }
            }
        }

        /**
         * Removes the given node, that must be present before this call.
         * This is messier than typical red-black deletion code because we
         * cannot swap the contents of an interior node with a leaf
         * successor that is pinned by "next" pointers that are accessible
         * independently during traversal. So instead we swap the tree
         * linkages. If the current tree appears to have too few nodes,
         * the bin is converted back to a plain bin. (The test triggers
         * somewhere between 2 and 6 nodes, depending on tree structure).
         */
        final void removeTreeNode(HashMap<K,V> map, Node<K,V>[] tab,
                                  boolean movable) {
            int n;
            if (tab == null || (n = tab.length) == 0)
                return;
            int index = (n - 1) & hash;
            TreeNode<K,V> first = (TreeNode<K,V>)tab[index], root = first, rl;
            TreeNode<K,V> succ = (TreeNode<K,V>)next, pred = prev;
            if (pred == null)
                tab[index] = first = succ;
            else
                pred.next = succ;
            if (succ != null)
                succ.prev = pred;
            if (first == null)
                return;
            if (root.parent != null)
                root = root.root();
            if (root == null || root.right == null ||
                (rl = root.left) == null || rl.left == null) {
                tab[index] = first.untreeify(map);  // too small
                return;
            }
            TreeNode<K,V> p = this, pl = left, pr = right, replacement;
            if (pl != null && pr != null) {
                TreeNode<K,V> s = pr, sl;
                while ((sl = s.left) != null) // find successor
                    s = sl;
                boolean c = s.red; s.red = p.red; p.red = c; // swap colors
                TreeNode<K,V> sr = s.right;
                TreeNode<K,V> pp = p.parent;
                if (s == pr) { // p was s's direct parent
                    p.parent = s;
                    s.right = p;
                }
                else {
                    TreeNode<K,V> sp = s.parent;
                    if ((p.parent = sp) != null) {
                        if (s == sp.left)
                            sp.left = p;
                        else
                            sp.right = p;
                    }
                    if ((s.right = pr) != null)
                        pr.parent = s;
                }
                p.left = null;
                if ((p.right = sr) != null)
                    sr.parent = p;
                if ((s.left = pl) != null)
                    pl.parent = s;
                if ((s.parent = pp) == null)
                    root = s;
                else if (p == pp.left)
                    pp.left = s;
                else
                    pp.right = s;
                if (sr != null)
                    replacement = sr;
                else
                    replacement = p;
            }
            else if (pl != null)
                replacement = pl;
            else if (pr != null)
                replacement = pr;
            else
                replacement = p;
            if (replacement != p) {
                TreeNode<K,V> pp = replacement.parent = p.parent;
                if (pp == null)
                    root = replacement;
                else if (p == pp.left)
                    pp.left = replacement;
                else
                    pp.right = replacement;
                p.left = p.right = p.parent = null;
            }

            TreeNode<K,V> r = p.red ? root : balanceDeletion(root, replacement);

            if (replacement == p) {  // detach
                TreeNode<K,V> pp = p.parent;
                p.parent = null;
                if (pp != null) {
                    if (p == pp.left)
                        pp.left = null;
                    else if (p == pp.right)
                        pp.right = null;
                }
            }
            if (movable)
                moveRootToFront(tab, r);
        }

        /* ------------------------------------------------------------ */
        // Red-black tree methods, all adapted from CLR
        //左旋
        static <K,V> TreeNode<K,V> rotateLeft(TreeNode<K,V> root,
                                              TreeNode<K,V> p) {
            TreeNode<K,V> r, pp, rl;
            //以p为左旋支点,且p不为空,右儿子不为空
            if (p != null && (r = p.right) != null) {
                //将p的右儿子r的左儿子rl变为p的右儿子
                if ((rl = p.right = r.left) != null)
                    rl.parent = p;
                //处理p、l和p父节点的关系
                if ((pp = r.parent = p.parent) == null)
                    (root = r).red = false;
                else if (pp.left == p)
                    pp.left = r;
                else
                    pp.right = r;
               //处理p和r的关系
                r.left = p;
                p.parent = r;
            }
            return root;
        }
        //右旋
        static <K,V> TreeNode<K,V> rotateRight(TreeNode<K,V> root,
                                               TreeNode<K,V> p) {
            TreeNode<K,V> l, pp, lr;
             //p:右旋支点,不为空,p的左儿子l不为空
            if (p != null && (l = p.left) != null) {
                 //将左儿子的右子树变为p的左子树
                if ((lr = p.left = l.right) != null)
                    lr.parent = p;
                 //p的父节点变为l的父节点
                if ((pp = l.parent = p.parent) == null)
                    (root = l).red = false;
             //如果p为右儿子,则p的父节点的右儿子变为l,否则左儿子变为l
                else if (pp.right == p)
                    pp.right = l;
                else
                    pp.left = l;
                //p变为l的右儿子
                l.right = p;
                p.parent = l;
            }
            return root;
        }

        static <K,V> TreeNode<K,V> balanceInsertion(TreeNode<K,V> root,
                                                    TreeNode<K,V> x) {
           //插入节点初始化为红色
            x.red = true;
            //xp:父节点,xpp:祖父节点, xppl:祖父节点的左儿子,xppr:祖父节点的右儿子
             //循环遍历
            for (TreeNode<K,V> xp, xpp, xppl, xppr;;) {
                 //插入的节点为根节点,节点颜色转换为黑色
                if ((xp = x.parent) == null) {
                    x.red = false;
                    return x;
                }
                 //当前节点的父为黑色节点或者父节点为根节点,直接返回
                else if (!xp.red || (xpp = xp.parent) == null)
                    return root;
                //祖父节点的左儿子是父节点
                if (xp == (xppl = xpp.left)) {
                    //S1:当前节点的父节点xp是红色,且当前节点的祖父节xpp点的另一个子节点(xppl或者xppr)也是红色
                    if ((xppr = xpp.right) != null && xppr.red) {
                        xppr.red = false;
                        xp.red = false;
                        xpp.red = true;
                        x = xpp;
                    }
                    else {
                         //S2:当前节点的父节点xp是红色,祖父节点的儿子节点(xppl或者xppr)是黑色,且当前节点x是其父节点xp的右孩子
                        if (x == xp.right) {
                            root = rotateLeft(root, x = xp);
                            xpp = (xp = x.parent) == null ? null : xp.parent;
                        }
                        //S3:当前节点的父节点xp是红色,祖父节点的儿子节点(xppl或者xppr)是黑色,且当前节点是其父节点xp的左孩子
                        if (xp != null) {
                            xp.red = false;
                            if (xpp != null) {
                                xpp.red = true;
                                root = rotateRight(root, xpp);
                            }
                        }
                    }
                }
                else {
                    //S1:当前节点的父节点xp是红色,且当前节点的祖父节xpp点的另一个子节点(xppl或者xppr)也是红色
                    if (xppl != null && xppl.red) {
                        xppl.red = false;
                        xp.red = false;
                        xpp.red = true;
                        x = xpp;
                    }
                    else {
                        //S2:当前节点的父节点xp是红色,祖父节点的儿子节点(xppl或者xppr)是黑色,且当前节点x是其父节点xp的右孩子
                        if (x == xp.left) {
                            root = rotateRight(root, x = xp);
                            xpp = (xp = x.parent) == null ? null : xp.parent;
                        }
                        //S3:当前节点的父节点xp是红色,祖父节点的儿子节点(xppl或者xppr)是黑色,且当前节点是其父节点xp的左孩子
                        if (xp != null) {
                            xp.red = false;
                            if (xpp != null) {
                                xpp.red = true;
                                root = rotateLeft(root, xpp);
                            }
                        }
                    }
                }
            }
        }

        static <K,V> TreeNode<K,V> balanceDeletion(TreeNode<K,V> root,
                                                   TreeNode<K,V> x) {
            for (TreeNode<K,V> xp, xpl, xpr;;)  {
                if (x == null || x == root)
                    return root;
                else if ((xp = x.parent) == null) {
                    x.red = false;
                    return x;
                }
                else if (x.red) {
                    x.red = false;
                    return root;
                }
                else if ((xpl = xp.left) == x) {
                    if ((xpr = xp.right) != null && xpr.red) {
                        xpr.red = false;
                        xp.red = true;
                        root = rotateLeft(root, xp);
                        xpr = (xp = x.parent) == null ? null : xp.right;
                    }
                    if (xpr == null)
                        x = xp;
                    else {
                        TreeNode<K,V> sl = xpr.left, sr = xpr.right;
                        if ((sr == null || !sr.red) &&
                            (sl == null || !sl.red)) {
                            xpr.red = true;
                            x = xp;
                        }
                        else {
                            if (sr == null || !sr.red) {
                                if (sl != null)
                                    sl.red = false;
                                xpr.red = true;
                                root = rotateRight(root, xpr);
                                xpr = (xp = x.parent) == null ?
                                    null : xp.right;
                            }
                            if (xpr != null) {
                                xpr.red = (xp == null) ? false : xp.red;
                                if ((sr = xpr.right) != null)
                                    sr.red = false;
                            }
                            if (xp != null) {
                                xp.red = false;
                                root = rotateLeft(root, xp);
                            }
                            x = root;
                        }
                    }
                }
                else { // symmetric
                    if (xpl != null && xpl.red) {
                        xpl.red = false;
                        xp.red = true;
                        root = rotateRight(root, xp);
                        xpl = (xp = x.parent) == null ? null : xp.left;
                    }
                    if (xpl == null)
                        x = xp;
                    else {
                        TreeNode<K,V> sl = xpl.left, sr = xpl.right;
                        if ((sl == null || !sl.red) &&
                            (sr == null || !sr.red)) {
                            xpl.red = true;
                            x = xp;
                        }
                        else {
                            if (sl == null || !sl.red) {
                                if (sr != null)
                                    sr.red = false;
                                xpl.red = true;
                                root = rotateLeft(root, xpl);
                                xpl = (xp = x.parent) == null ?
                                    null : xp.left;
                            }
                            if (xpl != null) {
                                xpl.red = (xp == null) ? false : xp.red;
                                if ((sl = xpl.left) != null)
                                    sl.red = false;
                            }
                            if (xp != null) {
                                xp.red = false;
                                root = rotateRight(root, xp);
                            }
                            x = root;
                        }
                    }
                }
            }
        }
  }

三、总结

    原文作者:算法小白
    原文地址: https://segmentfault.com/a/1190000016971486
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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