JDK 源码分析:TreeMap(二)

栏目: 编程工具 · 发布时间: 5年前

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&lt;K,V&gt; 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&lt;&gt;(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&lt;K,V&gt; <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>&lt;?</span> super K&gt; 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 &lt; <span>0</span>)</span>

<span> t = t.left;</span>

<span> <span>else</span> <span>if</span> (cmp &gt; <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>&quot;unchecked&quot;</span>)</span>

<span> Comparable<span>&lt;?</span> super K&gt; k = (Comparable<span>&lt;?</span> super K&gt;) 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 &lt; <span>0</span>)</span>

<span> t = t.left;</span>

<span> <span>else</span> <span>if</span> (cmp &gt; <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&lt;K,V&gt; e = <span>new</span> Entry&lt;&gt;(key, value, <span>parent</span>);</span>

<span> <span>if</span> (cmp &lt; <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

JDK 源码分析:TreeMap(二)

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

JDK 源码分析:TreeMap(二)

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

JDK 源码分析:TreeMap(二)

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&lt;K,V&gt; 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&lt;K,V&gt; 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&lt;K,V&gt; 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&lt;K,V&gt; 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&lt;K,V&gt; 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&lt;K,V&gt; 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

JDK 源码分析:TreeMap(二)

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

JDK 源码分析:TreeMap(二)

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

JDK 源码分析:TreeMap(二)

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

JDK 源码分析:TreeMap(二)

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:

JDK源码分析-TreeMap(1)

数据结构与算法笔记(四)

Stay hungry, stay foolish.

JDK 源码分析:TreeMap(二)


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