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我试图在Java中实现红黑树。为了做到这一点,我允许每个插入发生,就好像它是一个BST,然后后插入我的预订遍历树,并检查每个节点(我称为一个字,因为我用它字典应用程序)红黑规则得到满足。到目前为止,它看起来像这样Java中的红黑树规则实施
private void checkRedBlackRules(Word w)
{
//Checking for Red-Red sequence
Word leftChild = w.getLeft();
Word rightChild = w.getRight();
Word leftleftGC, leftrightGC, rightleftGC, rightrightGC;
if(leftChild != null)
{
leftleftGC = leftChild.getLeft();
leftrightGC = leftChild.getRight();
}
else
{
leftleftGC = null;
leftrightGC = null;
}
if(rightChild != null)
{
rightleftGC = rightChild.getLeft();
rightrightGC = rightChild.getRight();
}
else
{
rightleftGC = null;
rightrightGC = null;
}
try
{
if(leftChild.isRed() && leftleftGC.isRed())
{
rotateRight(w, leftChild, leftleftGC);
}
}
catch(NullPointerException e) {}
try
{
if(leftChild.isRed() && leftrightGC.isRed())
{
}
}
catch(NullPointerException e) {}
try
{
if(rightChild.isRed() && rightleftGC.isRed())
{
}
}
catch(NullPointerException e) {}
try
{
if(rightChild.isRed() && rightrightGC.isRed())
{
rotateLeft(w, leftChild, leftrightGC);
}
}
catch(NullPointerException e) {}
if(w.getLeft() != null)
checkRedBlackRules(w.getLeft());
if(w.getRight() != null)
checkRedBlackRules(w.getRight());
}
private void rotateLeft(Word parent, Word child, Word grandChild)
{
child = parent;
child.setLeft(parent);
child.setRight(grandChild);
}
private void rotateRight(Word parent, Word child, Word grandChild)
{
child = parent;
child.setLeft(grandChild);
child.setRight(parent);
}
然而,当我尝试两个以上的元素添加到它陷在一个无限循环,直到发生StackOverflow的错误树。
为什么你不分青红皂白地捕捉和忽略'NullPointerException's?这是_definitely_不是正确的做法。 – 2015-03-18 23:16:14
旋转函数看起来很奇怪......'child = parent; child.setLeft(parent);'就像说'parent.setLeft(parent);'。 – Alan 2015-03-18 23:18:59