内容简介:前文「插入操作该操作其实就是红黑树的插入节点操作。
前文「 JDK源码分析-TreeMap(1) 」分析了 TreeMap 的一些方法,本文分析其中的增删方法。这也是红黑树插入和删除节点的操作,由于相对复杂,因此单独进行分析。
插入操作
该操作其实就是红黑树的插入节点操作。 前面分析过, 红黑树是一种平衡二叉树,新增 节点后 可能导致其失去平衡,因此需要对其进行修复操作以维持其平衡性。插入操作的 代码如下:
<span><span>public</span> V put(K key, V value) {</span>
<span> Entry<K,V> t = root;</span>
<span> <span>// 若 root 节点为空,则直接插入(为根节点)</span></span>
<span> <span>if</span> (t == <span>null</span>) {</span>
<span> compare(key, key); <span>// type (and possibly null) check</span></span>
<span> root = <span>new</span> Entry<>(key, value, <span>null</span>);</span>
<span> size = <span>1</span>;</span>
<span> modCount++;</span>
<span> <span>return</span> <span>null</span>;</span>
<span> }</span>
<span> int cmp;</span>
<span> Entry<K,V> <span>parent</span>;</span>
<span> <span>// split comparator and comparable paths</span></span>
<span> <span>// 拆分 Comparator 接口和 Comparable 接口(上文 getEntry 方法也是如此)</span></span>
<span> Comparator<span><?</span> super K> cpr = comparator;</span>
<span> <span>if</span> (cpr != <span>null</span>) {</span>
<span> <span>do</span> {</span>
<span> <span>parent</span> = t;</span>
<span> cmp = cpr.compare(key, t.key);</span>
<span> <span>if</span> (cmp < <span>0</span>)</span>
<span> t = t.left;</span>
<span> <span>else</span> <span>if</span> (cmp > <span>0</span>)</span>
<span> t = t.right;</span>
<span> <span>else</span></span>
<span> <span>// 若key已存在,则替换其对应的value</span></span>
<span> <span>return</span> t.setValue(value);</span>
<span> } <span>while</span> (t != <span>null</span>);</span>
<span> }</span>
<span> <span>else</span> {</span>
<span> <span>if</span> (key == <span>null</span>)</span>
<span> <span>throw</span> <span>new</span> NullPointerException();</span>
<span> @SuppressWarnings(<span>"unchecked"</span>)</span>
<span> Comparable<span><?</span> super K> k = (Comparable<span><?</span> super K>) key;</span>
<span> <span>do</span> {</span>
<span> <span>parent</span> = t;</span>
<span> cmp = k.compareTo(t.key);</span>
<span> <span>if</span> (cmp < <span>0</span>)</span>
<span> t = t.left;</span>
<span> <span>else</span> <span>if</span> (cmp > <span>0</span>)</span>
<span> t = t.right;</span>
<span> <span>else</span></span>
<span> <span>return</span> t.setValue(value);</span>
<span> } <span>while</span> (t != <span>null</span>);</span>
<span> }</span>
<span> Entry<K,V> e = <span>new</span> Entry<>(key, value, <span>parent</span>);</span>
<span> <span>if</span> (cmp < <span>0</span>)</span>
<span> <span>parent</span>.left = e;</span>
<span> <span>else</span></span>
<span> <span>parent</span>.right = e;</span>
<span> <span>// 插入节点后的平衡性调整</span></span>
<span> fixAfterInsertion(e);</span>
<span> size++;</span>
<span> modCount++;</span>
<span> <span>return</span> <span>null</span>;</span>
<span>}</span>
对应的几种插入节点修复操作前文「 数据结构与算法笔记(四) 」 已进行了分析 ,为了便于分析和理解代码,这里把图再贴一下(下图为关注节点的父节点是其祖父节点的左子节点的情况,在右边时操作类似):
case1: 关注节点 a 的叔叔节点为红色
case2: 关注节点为 a, 它的叔叔节点 d 是黑色,a 是其父节点 b 的右子节点
case3: 关注节点是 a,它的叔叔节点 d 是黑色,a 是其父节点 b 的左子节点
插入操作的平衡调整代码如下:
<span><span><span>private</span> <span>void</span> <span>fixAfterInsertion</span>(<span>Entry<K,V> x</span>)</span> {</span>
<span> <span>// 新插入的节点为红色</span></span>
<span> x.color = RED;</span>
<span> <span>// 只有在父节点为红色时需要进行插入修复操作</span></span>
<span> <span>while</span> (x != <span>null</span> && x != root && x.parent.color == RED) {</span>
<span> <span>// 下面两种情况是左右对称的</span></span>
<span> <span>// x 的父节点是它祖父节点的左子节点</span></span>
<span> <span>if</span> (parentOf(x) == leftOf(parentOf(parentOf(x)))) {</span>
<span> <span>// 叔叔节点</span></span>
<span> Entry<K,V> y = rightOf(parentOf(parentOf(x)));</span>
<span> <span>// case1</span></span>
<span> <span>if</span> (colorOf(y) == RED) {</span>
<span> setColor(parentOf(x), BLACK);</span>
<span> setColor(y, BLACK);</span>
<span> setColor(parentOf(parentOf(x)), RED);</span>
<span> x = parentOf(parentOf(x));</span>
<span> } <span>else</span> {</span>
<span> <span>// case2</span></span>
<span> <span>if</span> (x == rightOf(parentOf(x))) {</span>
<span> x = parentOf(x);</span>
<span> rotateLeft(x);</span>
<span> }</span>
<span> <span>// case3</span></span>
<span> setColor(parentOf(x), BLACK);</span>
<span> setColor(parentOf(parentOf(x)), RED);</span>
<span> rotateRight(parentOf(parentOf(x)));</span>
<span> }</span>
<span> } </span>
<span> <span>// x 的父节点是它祖父节点的右子节点(与上面情况对称)</span></span>
<span> <span>else</span> {</span>
<span> Entry<K,V> y = leftOf(parentOf(parentOf(x)));</span>
<span> <span>if</span> (colorOf(y) == RED) {</span>
<span> setColor(parentOf(x), BLACK);</span>
<span> setColor(y, BLACK);</span>
<span> setColor(parentOf(parentOf(x)), RED);</span>
<span> x = parentOf(parentOf(x));</span>
<span> } <span>else</span> {</span>
<span> <span>if</span> (x == leftOf(parentOf(x))) {</span>
<span> x = parentOf(x);</span>
<span> rotateRight(x);</span>
<span> }</span>
<span> setColor(parentOf(x), BLACK);</span>
<span> setColor(parentOf(parentOf(x)), RED);</span>
<span> rotateLeft(parentOf(parentOf(x)));</span>
<span> }</span>
<span> }</span>
<span> }</span>
<span> root.color = BLACK;</span>
<span>}</span>
对称情况下的相应操作不再分析,其原理是类似的。
删除操作
remove() 方法:
public V remove(Object key) {
Entry<K,V> p = getEntry(key);
if (p == null)
return null;
V oldValue = p.value;
deleteEntry(p);
return oldValue;
}
内部实现方法如下:
<span><span>/**</span></span>
<span><span> * Delete node p, and then rebalance the tree.</span></span>
<span><span> */</span></span>
<span><span>private</span> void deleteEntry(Entry<K,V> p) {</span>
<span> modCount++;</span>
<span> size--;</span>
<span> <span>// If strictly internal, copy successor's element to p and then make p</span></span>
<span> <span>// point to successor.</span></span>
<span> <span>// 左右子树都不为空,寻找后继节点</span></span>
<span> <span>if</span> (p.left != <span>null</span> && p.right != <span>null</span>) {</span>
<span> Entry<K,V> s = successor(p);</span>
<span> p.key = s.key;</span>
<span> p.value = s.value;</span>
<span> p = s;</span>
<span> } <span>// p has 2 children</span></span>
<span> <span>// Start fixup at replacement node, if it exists.</span></span>
<span> Entry<K,V> replacement = (p.left != <span>null</span> ? p.left : p.right);</span>
<span> <span>if</span> (replacement != <span>null</span>) {</span>
<span> <span>// Link replacement to parent</span></span>
<span> replacement.<span>parent</span> = p.<span>parent</span>;</span>
<span> <span>if</span> (p.<span>parent</span> == <span>null</span>)</span>
<span> root = replacement;</span>
<span> <span>else</span> <span>if</span> (p == p.<span>parent</span>.left)</span>
<span> p.<span>parent</span>.left = replacement;</span>
<span> <span>else</span></span>
<span> p.<span>parent</span>.right = replacement;</span>
<span> <span>// Null out links so they are OK to use by fixAfterDeletion.</span></span>
<span> p.left = p.right = p.<span>parent</span> = <span>null</span>;</span>
<span> <span>// Fix replacement</span></span>
<span> <span>if</span> (p.color == BLACK)</span>
<span> fixAfterDeletion(replacement);</span>
<span> } <span>else</span> <span>if</span> (p.<span>parent</span> == <span>null</span>) { <span>// return if we are the only node.</span></span>
<span> <span>// 只有一个根节点</span></span>
<span> root = <span>null</span>;</span>
<span> } <span>else</span> { <span>// No children. Use self as phantom replacement and unlink.</span></span>
<span> <span>if</span> (p.color == BLACK)</span>
<span> fixAfterDeletion(p);</span>
<span> <span>if</span> (p.<span>parent</span> != <span>null</span>) {</span>
<span> <span>if</span> (p == p.<span>parent</span>.left)</span>
<span> p.<span>parent</span>.left = <span>null</span>;</span>
<span> <span>else</span> <span>if</span> (p == p.<span>parent</span>.right)</span>
<span> p.<span>parent</span>.right = <span>null</span>;</span>
<span> p.<span>parent</span> = <span>null</span>;</span>
<span> }</span>
<span> }</span>
<span>}</span>
几种删除操作情况如下(下图为关注节点为父节点的左子节点的情况,关注节点为父节点的右子节点情况时的操作对称):
case1: 关注 节点的兄弟节点是红色
case2: 关注 节点的兄弟节点是黑色,且兄弟节点的子节点都是黑色
case3: 关注 节点的兄弟节点是黑色,且左子节点是红色、右子节点是黑色
case4: 关注 节点的兄弟节点是黑色,且右子节点是红色、左子节点是黑色
勘误 :前文「 数据结构与算法笔记(四) 」对红黑树删除操作第四种情况的分析不够准确,近两天又参考了其他文章及代码,这里的 case4 是目前经分析认为比较准确的(符合 JDK 1.8 源码中 TreeMap 的实现思路)。
PS: 别人的资料也未必都正确,不可全信,包括本文,还是要持有怀疑精神的。
删除操作的平衡调整代码如下:
private void fixAfterDeletion(Entry<K,V> x) {
// x 不为根节点,且颜色为黑色
while (x != root && colorOf(x) == BLACK) {
// x 是父节点的左子节点
if (x == leftOf(parentOf(x))) {
// 兄弟节点
Entry<K,V> sib = rightOf(parentOf(x));
// case1 待删除节点的兄弟节点为红色
if (colorOf(sib) == RED) {
setColor(sib, BLACK);
setColor(parentOf(x), RED);
rotateLeft(parentOf(x));
sib = rightOf(parentOf(x));
}
// case2 待删除节点的兄弟节点的子节点都为黑色
if (colorOf(leftOf(sib)) == BLACK && colorOf(rightOf(sib)) == BLACK) {
setColor(sib, RED);
x = parentOf(x);
} else {
// case3 待删除节点的兄弟节点的左子节点为红色、右子节为黑色
if (colorOf(rightOf(sib)) == BLACK) {
setColor(leftOf(sib), BLACK);
setColor(sib, RED);
rotateRight(sib);
sib = rightOf(parentOf(x));
}
// case4 待删除节点的兄弟节点的左子节点为黑色、右子节为红色
setColor(sib, colorOf(parentOf(x)));
setColor(parentOf(x), BLACK);
setColor(rightOf(sib), BLACK); //??
rotateLeft(parentOf(x));
x = root;
}
}
// x 是父节点的右子节点(对称操作)
else { // symmetric
Entry<K,V> sib = leftOf(parentOf(x));
if (colorOf(sib) == RED) {
setColor(sib, BLACK);
setColor(parentOf(x), RED);
rotateRight(parentOf(x));
sib = leftOf(parentOf(x));
}
if (colorOf(rightOf(sib)) == BLACK &&
colorOf(leftOf(sib)) == BLACK) {
setColor(sib, RED);
x = parentOf(x);
} else {
if (colorOf(leftOf(sib)) == BLACK) {
setColor(rightOf(sib), BLACK);
setColor(sib, RED);
rotateLeft(sib);
sib = leftOf(parentOf(x));
}
setColor(sib, colorOf(parentOf(x)));
setColor(parentOf(x), BLACK);
setColor(leftOf(sib), BLACK);
rotateRight(parentOf(x));
x = root;
}
}
}
setColor(x, BLACK);
}
插入和删除操作相对复杂,容易被绕晕,但其实也是有规律可循的。对比操作的图解,可以更容易分析和理解。
参考文章:
https: //zhuanlan.zhihu.com/p/22800206
这篇文章介绍了红黑树的删除操作,逻辑清晰,推荐阅读。
相关阅读:
Stay hungry, stay foolish.
以上所述就是小编给大家介绍的《JDK 源码分析:TreeMap(二)》,希望对大家有所帮助,如果大家有任何疑问请给我留言,小编会及时回复大家的。在此也非常感谢大家对 码农网 的支持!
猜你喜欢:- 以太坊源码分析(36)ethdb源码分析
- [源码分析] kubelet源码分析(一)之 NewKubeletCommand
- libmodbus源码分析(3)从机(服务端)功能源码分析
- [源码分析] nfs-client-provisioner源码分析
- [源码分析] kubelet源码分析(三)之 Pod的创建
- Spring事务源码分析专题(一)JdbcTemplate使用及源码分析
本站部分资源来源于网络,本站转载出于传递更多信息之目的,版权归原作者或者来源机构所有,如转载稿涉及版权问题,请联系我们。
我看电商2(双色)
黄若 / 电子工业出版社 / 2016-6 / 39.00元
《我看电商2》是行业畅销书《我看电商》的续集。 《我看电商》自出版以来,连续印刷14 次,受到业界人士和广大读者的高度好评。《我看电商2》承续作者一贯的风格,以行业观察、经验分享为出发点,重点分析了过去一年中国电商界的最新动态与趋势,包括双11点评、京东关闭拍拍、上市公司私有化等。 电子商务是我国近年来发展最快的新兴行业之一,作者作为这个行业的长老级领军人物,善于思考,长于实操。《我看......一起来看看 《我看电商2(双色)》 这本书的介绍吧!