OpenJDK 源代码阅读之 TreeMap
概要
- 类继承关系
java.lang.Object
java.util.AbstractMap<K,V>
java.util.HashMap<K,V>
- 定义
public class TreeMap<K,V>
extends AbstractMap<K,V>
implements NavigableMap<K,V>, Cloneable, java.io.Serializable
- 要点
1) 基于 NavigableMap 实现的红黑树
2) 按 natrual ordering 或者 Comparator 定义的次序排序。
3) 基本操作 containsKey,get,put 有 log(n) 的时间复杂度。
4) 非线程安全
实现
- Comparator
private final Comparator<? super K> comparator;
这说明提供的 Comparator 参数类型是 K 的基类就行。这似乎意味着基类的 Comparator 与导出类的要一致。
- Entry
private transient Entry<K,V> root = null;
root 是这棵红黑树的根，那么从 Entry 的定义可以体现树的结构：
static final class Entry<K,V> implements Map.Entry<K,V> {
K key;
V value;
Entry<K,V> left = null;
Entry<K,V> right = null;
Entry<K,V> parent;
boolean color = BLACK;
...
}
注意这是个 static final 类，left,right,parent 分别指向左子树，右子树，父结点, color 颜色默认为黑。
- containsKey
public boolean containsKey(Object key) {
return getEntry(key) != null;
}
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();
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;
}
关键操作 containsKey 是通过调用 getEntry 完成其功能的。可以看出，这是通过在红黑树上进行查找完成的，每次比较都会下降到树的下一层，由于红黑树的平衡性，时间复杂度为 log(n)。
- containsValue
public boolean containsValue(Object value) {
for (Entry<K,V> e = getFirstEntry(); e != null; e = successor(e))
if (valEquals(value, e.value))
return true;
return false;
}
可以看出，对 value 的查找，与对 key 是不同的。是通过 getFirstEntry 取得第一个结点，再通过 successor 遍历实现。
final Entry<K,V> getFirstEntry() {
Entry<K,V> p = root;
if (p != null)
while (p.left != null)
p = p.left;
return p;
}
可以看出，这是找到了树中最左边的结点，如果左子树中的值小于右子树，这就意味是第一个比较的结点是最小的结点。
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;
}
}
successor 其实就是一个结点的 下一个结点，所谓 下一个，是按次序排序后的下一个结点。从代码中可以看出，如果右子树不为空，就返回右子树中最小结点。如果右子树为空，就要向上回溯了。在这种情况下，t 是以其为根的树的最后一个结点。如果它是其父结点的左孩子，那么父结点就是它的下一个结点，否则，t 就是以其父结点为根的树的最后一个结点，需要再次向上回溯。一直到 ch 是 p 的左孩子为止。
- getCeilingEntry
/**
* Gets the entry corresponding to the specified key; if no such entry
* exists, returns the entry for the least key greater than the specified
* key; if no such entry exists (i.e., the greatest key in the Tree is less
* than the specified key), returns {@code null}.
*/
final Entry<K,V> getCeilingEntry(K key) {
Entry<K,V> p = root;
while (p != null) {
int cmp = compare(key, p.key);
if (cmp < 0) {
if (p.left != null)
p = p.left;
else
return p;
} else if (cmp > 0) {
if (p.right != null) {
p = p.right;
} else {
Entry<K,V> parent = p.parent;
Entry<K,V> ch = p;
while (parent != null && ch == parent.right) {
ch = parent;
parent = parent.parent;
}
return parent;
}
} else
return p;
}
return null;
}
这个函数看起来有点奇怪，可以从 return 语句，猜想一下这是在做什么，我觉得是在找一个 x.key >= key 的元素，并且 x 是满足条件的元素中最小的。从 23 行看，找到的 p，key 值与参数 key 相等，从 8 行看，又有 key <= p.key，并且 p.left == null。
put
/**
* 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();
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;
}
put 的过程，其实是将 (key, value) 加入到红黑树中的过程。如果树是空的，那么创建根结点。否则就要在树中插入结点。这个过程根据 comparator 是否存在设置成了两种方式，其实没什么区别，就是比较方式的不同，都是在树中查找一个合适的位置，如果 key 在树中，就 setValue 设置新值，否则，就在 60-64 行插入新结点。这里有个重要的地方是 65 行的 fixAfterInsertion ，这个很重要，因为这是一棵红黑树，红黑树关键的思想是要保持它的平衡性，插入结点后，平衡性可能被破坏。所以需要 fix。
/** From CLR */
private void fixAfterInsertion(Entry<K,V> x) {
x.color = RED;
while (x != null && x != root && x.parent.color == RED) {
if (parentOf(x) == leftOf(parentOf(parentOf(x)))) {
Entry<K,V> y = rightOf(parentOf(parentOf(x)));
if (colorOf(y) == RED) {
setColor(parentOf(x), BLACK);
setColor(y, BLACK);
setColor(parentOf(parentOf(x)), RED);
x = parentOf(parentOf(x));
} else {
if (x == rightOf(parentOf(x))) {
x = parentOf(x);
rotateLeft(x);
}
setColor(parentOf(x), BLACK);
setColor(parentOf(parentOf(x)), RED);
rotateRight(parentOf(parentOf(x)));
}
} else {
Entry<K,V> y = leftOf(parentOf(parentOf(x)));
if (colorOf(y) == RED) {
setColor(parentOf(x), BLACK);
setColor(y, BLACK);
setColor(parentOf(parentOf(x)), RED);
x = parentOf(parentOf(x));
} else {
if (x == leftOf(parentOf(x))) {
x = parentOf(x);
rotateRight(x);
}
setColor(parentOf(x), BLACK);
setColor(parentOf(parentOf(x)), RED);
rotateLeft(parentOf(parentOf(x)));
}
}
}
root.color = BLACK;
}
这是 fixAfterInsertion 的源代码，我不具体解释了，这是一个根据不同情况进行旋转，调整结点颜色的过程，可以参考《算法导论》中的解释。
- remove
public V remove(Object key) {
Entry<K,V> p = getEntry(key);
if (p == null)
return null;
V oldValue = p.value;
deleteEntry(p);
return oldValue;
}
删除的过程主要调用 delteEntry 完成：
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;
}
}
}
这个过程同样是与红黑树的性质相关的。在树中删除一个结点，那他的孩子怎么办啊，红黑树不平衡了怎么办啊，树空了怎么办啊，都需要考虑到。
- clear
public void clear() {
modCount++;
size = 0;
root = null;
}
居然就是把值都清空了。。
- clone
/**
* Returns a shallow copy of this {@code TreeMap} instance. (The keys and
* values themselves are not cloned.)
*
* @return a shallow copy of this map
*/
public Object clone() {
TreeMap<K,V> clone = null;
try {
clone = (TreeMap<K,V>) super.clone();
} catch (CloneNotSupportedException e) {
throw new InternalError();
}
// Put clone into "virgin" state (except for comparator)
clone.root = null;
clone.size = 0;
clone.modCount = 0;
clone.entrySet = null;
clone.navigableKeySet = null;
clone.descendingMap = null;
// Initialize clone with our mappings
try {
clone.buildFromSorted(size, entrySet().iterator(), null, null);
} catch (java.io.IOException cannotHappen) {
} catch (ClassNotFoundException cannotHappen) {
}
return clone;
}
clone 复制了一份新的元素，使用了 super.clone() 得的一个新对象，而不是使用 new，这是为啥？然后把各个域值清空，然后使用 buildFromSorted 插入数据。之所以使用这个函数，是因为红黑树插入操作时间复杂度为 O(lgn)，n个元素插入就是 O(n*lgn)，太不划算，更何况我们现在插入的是一个红黑树，所以用一个线性时间复杂度的算法来实现复制数据的操作。
/**
* Linear time tree building algorithm from sorted data. Can accept keys
* and/or values from iterator or stream. This leads to too many
* parameters, but seems better than alternatives. The four formats
* that this method accepts are:
*
* 1) An iterator of Map.Entries. (it != null, defaultVal == null).
* 2) An iterator of keys. (it != null, defaultVal != null).
* 3) A stream of alternating serialized keys and values.
* (it == null, defaultVal == null).
* 4) A stream of serialized keys. (it == null, defaultVal != null).
*
* It is assumed that the comparator of the TreeMap is already set prior
* to calling this method.
*
* @param size the number of keys (or key-value pairs) to be read from
* the iterator or stream
* @param it If non-null, new entries are created from entries
* or keys read from this iterator.
* @param str If non-null, new entries are created from keys and
* possibly values read from this stream in serialized form.
* Exactly one of it and str should be non-null.
* @param defaultVal if non-null, this default value is used for
* each value in the map. If null, each value is read from
* iterator or stream, as described above.
* @throws IOException propagated from stream reads. This cannot
* occur if str is null.
* @throws ClassNotFoundException propagated from readObject.
* This cannot occur if str is null.
*/
private void buildFromSorted(int size, Iterator it,
java.io.ObjectInputStream str,
V defaultVal)
throws java.io.IOException, ClassNotFoundException {
this.size = size;
root = buildFromSorted(0, 0, size-1, computeRedLevel(size),
it, str, defaultVal);
}
好吧，我们继续。
private final Entry<K,V> buildFromSorted(int level, int lo, int hi,
int redLevel,
Iterator it,
java.io.ObjectInputStream str,
V defaultVal)
throws java.io.IOException, ClassNotFoundException {
/*
* Strategy: The root is the middlemost element. To get to it, we
* have to first recursively construct the entire left subtree,
* so as to grab all of its elements. We can then proceed with right
* subtree.
*
* The lo and hi arguments are the minimum and maximum
* indices to pull out of the iterator or stream for current subtree.
* They are not actually indexed, we just proceed sequentially,
* ensuring that items are extracted in corresponding order.
*/
if (hi < lo) return null;
int mid = (lo + hi) >>> 1;
Entry<K,V> left = null;
if (lo < mid)
left = buildFromSorted(level+1, lo, mid - 1, redLevel,
it, str, defaultVal);
// extract key and/or value from iterator or stream
K key;
V value;
if (it != null) {
if (defaultVal==null) {
Map.Entry<K,V> entry = (Map.Entry<K,V>)it.next();
key = entry.getKey();
value = entry.getValue();
} else {
key = (K)it.next();
value = defaultVal;
}
} else { // use stream
key = (K) str.readObject();
value = (defaultVal != null ? defaultVal : (V) str.readObject());
}
Entry<K,V> middle = new Entry<>(key, value, null);
// color nodes in non-full bottommost level red
if (level == redLevel)
middle.color = RED;
if (left != null) {
middle.left = left;
left.parent = middle;
}
if (mid < hi) {
Entry<K,V> right = buildFromSorted(level+1, mid+1, hi, redLevel,
it, str, defaultVal);
middle.right = right;
right.parent = middle;
}
return middle;
}
可以看出这是一个递归的算法，找出中间结点，构造左右子树，并将他们连接在一起。时间复杂度分析可以根据递归式：
T(n) = 2 * T(n/2) + C
如果看不出来的话，可以使用《算法导论》第4章中的主定理，可以得到时间复杂度为 O(n) 。