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源码看JAVA【二十五】TreeMap

程序员文章站 2024-03-18 08:15:40
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1、内部类Entry定义树键值对的结构

键值对的特征:key-value

tree的特征:left-right-parent

color:红黑树的设定

​
    private static final boolean RED   = false;
    private static final boolean BLACK = true;
    /**
     * Node in the Tree.  Doubles as a means to pass key-value pairs back to
     * user (see Map.Entry).
     */

    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;

        /**
         * Make a new cell with given key, value, and parent, and with
         * {@code null} child links, and BLACK color.
         */
        Entry(K key, V value, Entry<K,V> parent) {
            this.key = key;
            this.value = value;
            this.parent = parent;
        }

        /**
         * Returns the key.
         *
         * @return the key
         */
        public K getKey() {
            return key;
        }

        /**
         * Returns the value associated with the key.
         *
         * @return the value associated with the key
         */
        public V getValue() {
            return value;
        }

        /**
         * Replaces the value currently associated with the key with the given
         * value.
         *
         * @return the value associated with the key before this method was
         *         called
         */
        public V setValue(V value) {
            V oldValue = this.value;
            this.value = value;
            return oldValue;
        }

        public boolean equals(Object o) {
            if (!(o instanceof Map.Entry))
                return false;
            Map.Entry<?,?> e = (Map.Entry<?,?>)o;

            return valEquals(key,e.getKey()) && valEquals(value,e.getValue());
        }

        public int hashCode() {
            int keyHash = (key==null ? 0 : key.hashCode());
            int valueHash = (value==null ? 0 : value.hashCode());
            return keyHash ^ valueHash;
        }

        public String toString() {
            return key + "=" + value;
        }
    }

​

2、get(Object key)

进行比较判断,左边的值永远比右边的值大,根据此方式一次往下搜索。key值不能为空

    public V get(Object key) {
        Entry<K,V> p = getEntry(key);
        return (p==null ? null : p.value);
    }
    final Entry<K,V> getEntry(Object key) {
        // Offload comparator-based version for sake of performance
        if (comparator != null)
            return getEntryUsingComparator(key);
        if (key == null)
            throw new NullPointerException();
        @SuppressWarnings("unchecked")
            Comparable<? super K> k = (Comparable<? super K>) key;
        Entry<K,V> p = root;
        while (p != null) {
            int cmp = k.compareTo(p.key);
            if (cmp < 0)
                p = p.left;
            else if (cmp > 0)
                p = p.right;
            else
                return p;
        }
        return null;
    }

3、put(K key, V value):红黑树的添加元素

a、判断是不是首个节点

b、比较每个节点key的值,一直向下比较,直到节点空为止,或者key存在,直接设置值返回。

c、得到空节点之后,创建当前空节点。

d、因为保存了空节点的上个节点,所以得到了父节点,将新节点按比较的值挂到父节点的左值(新<父)/右值(新>=旧)

e、插入节点后,重构红黑树。重构方式可参考红黑树的构造逻辑。

    /**
     * Associates the specified value with the specified key in this map.
     * If the map previously contained a mapping for the key, the old
     * value is replaced.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     *
     * @return the previous value associated with {@code key}, or
     *         {@code null} if there was no mapping for {@code key}.
     *         (A {@code null} return can also indicate that the map
     *         previously associated {@code null} with {@code key}.)
     * @throws ClassCastException if the specified key cannot be compared
     *         with the keys currently in the map
     * @throws NullPointerException if the specified key is null
     *         and this map uses natural ordering, or its comparator
     *         does not permit null keys
     */
    public V put(K key, V value) {
        Entry<K,V> t = root;
        if (t == null) {
            compare(key, key); // type (and possibly null) check

            root = new Entry<>(key, value, null);
            size = 1;
            modCount++;
            return null;
        }
        int cmp;
        Entry<K,V> parent;
        // split comparator and comparable paths
        Comparator<? super K> cpr = comparator;
        if (cpr != null) {
            do {
                parent = t;
                cmp = cpr.compare(key, t.key);
                if (cmp < 0)
                    t = t.left;
                else if (cmp > 0)
                    t = t.right;
                else
                    return t.setValue(value);
            } while (t != null);
        }
        else {
            if (key == null)
                throw new NullPointerException();
            @SuppressWarnings("unchecked")
                Comparable<? super K> k = (Comparable<? super K>) key;
            do {
                parent = t;
                cmp = k.compareTo(t.key);
                if (cmp < 0)
                    t = t.left;
                else if (cmp > 0)
                    t = t.right;
                else
                    return t.setValue(value);
            } while (t != null);
        }
        Entry<K,V> e = new Entry<>(key, value, parent);
        if (cmp < 0)
            parent.left = e;
        else
            parent.right = e;
        fixAfterInsertion(e);
        size++;
        modCount++;
        return null;
    }

4、remove(Object key):删除节点

a、如果存在左节点且存在右节点,则将当前节点降级,规则为左节点的最右边节点或者右节点的最左边节点。

successor:找到节点

p.key = s.key;
p.value = s.value;

将找到的节点设置到当前节点,这样就可以删除找到的节点。找到的节点肯定最多只有一个左节点或者一个右节点。

b、如果左节点不为空,则将左节点替换到当前的节点。如果右节点不为空,则右节点替换到当前的节点。如果两个都为空,则判断父节点是否为空,为空则说明当前节点为根节点。如果左右节点都为空,则设置上级节点的子节点为空。把当前节点的父节点设置为空。

c、节点删除之后重新整理红黑树。

 

 

    public V remove(Object key) {
        Entry<K,V> p = getEntry(key);
        if (p == null)
            return null;

        V oldValue = p.value;
        deleteEntry(p);
        return oldValue;
    }

    /**
     * Delete node p, and then rebalance the tree.
     */
    private void deleteEntry(Entry<K,V> p) {
        modCount++;
        size--;

        // If strictly internal, copy successor's element to p and then make p
        // point to successor.
        if (p.left != null && p.right != null) {
            Entry<K,V> s = successor(p);
            p.key = s.key;
            p.value = s.value;
            p = s;
        } // p has 2 children

        // Start fixup at replacement node, if it exists.
        Entry<K,V> replacement = (p.left != null ? p.left : p.right);

        if (replacement != null) {
            // Link replacement to parent
            replacement.parent = p.parent;
            if (p.parent == null)
                root = replacement;
            else if (p == p.parent.left)
                p.parent.left  = replacement;
            else
                p.parent.right = replacement;

            // Null out links so they are OK to use by fixAfterDeletion.
            p.left = p.right = p.parent = null;

            // Fix replacement
            if (p.color == BLACK)
                fixAfterDeletion(replacement);
        } else if (p.parent == null) { // return if we are the only node.
            root = null;
        } else { //  No children. Use self as phantom replacement and unlink.
            if (p.color == BLACK)
                fixAfterDeletion(p);

            if (p.parent != null) {
                if (p == p.parent.left)
                    p.parent.left = null;
                else if (p == p.parent.right)
                    p.parent.right = null;
                p.parent = null;
            }
        }
    }

    /**
     * Returns the successor of the specified Entry, or null if no such.
     */
    static <K,V> TreeMap.Entry<K,V> successor(Entry<K,V> t) {
        if (t == null)
            return null;
        else if (t.right != null) {
            Entry<K,V> p = t.right;
            while (p.left != null)
                p = p.left;
            return p;
        } else {
            Entry<K,V> p = t.parent;
            Entry<K,V> ch = t;
            while (p != null && ch == p.right) {
                ch = p;
                p = p.parent;
            }
            return p;
        }
    }