我在这个家庭作业的插入方法上遇到了一些困难。我已经完成了大部分工作，但出于某种原因，每当我的程序应该在树的左侧插入一个节点作为右子项时，它就会将其作为左子项插入。
我以一种奇怪的方式进行比较（对于很多人来说，标志应该是相反的，但是由于某种原因，它的作用就是这样）所以如果你难以阅读它，请耐心等待。
我知道这是实现二叉搜索树的一种可怕方式，我永远不会在现实世界中做到这一点，但它是家庭作业，因此 - 我别无选择。
任何和所有帮助都一如既往地受到赞赏。谢谢！
编辑：我知道问题出在哪里。它在searchFor（）方法中。它不是节点的合法父节点，而是使父节点成为树的根节点（在这种情况下，节点的父节点始终是“cup”。）
现在已经不在了，任何人都可以提供解决方案吗？
Edit2：拿出了一些我认为与问题不相关的额外内容。我很确定我已将其缩小到searchFor（）方法。每当我调用返回当前节点的父节点时，它将返回树的根（“cup”。）我认为这就是我遇到问题的原因，因为它基于此插入。
感谢迄今为止的所有帮助，我真的很感激。
public class BinarySearchTree //implements Comparator
{
private Comparator<Object> dataComparator;
private LinkedListWithTwoLinks tree;
public static void main (String[] args)
{
BinarySearchTree bst;
Object hold;
String[] words = {"cup", "shaker", "cord", "key", "addressbook", "date", "address", "cupcake",
"card", "tape", "page", "day", "key", "days", "dayt"};
bst = new BinarySearchTree(new AlphabeticComparator());
System.out.println("[1]: original tree");
for(int i=0; i<words.length; i++) if (!bst.insert(words[i])) { System.out.println(">>>>>>>>>>>>> " + words[i] + " is already in tree"); }
bst.inOrder();
}
public static class AlphabeticComparator implements Comparator <Object>
{
public int compare(Object x, Object y)
{
if ( x == y ) return 0;
if ( x == null) return -1;
if ( y == null) return 1;
return (x.toString().compareTo(y.toString()));
}
}
public static class LastCharacterComparator implements Comparator <Object>
{
public int compare(Object x, Object y)
{
String xs;
String ys;
if ( x == y ) return 0;
if ( x == null ) return -1;
if ( y == null) return 1;
xs = x.toString();
ys = y.toString();
if ( xs.length() == 0) return -1;
if ( ys.length() == 0) return 1;
return (xs.charAt(xs.length()-1) - ys.charAt(ys.length()-1));
}
}
public BinarySearchTree(Comparator<Object> y)
{
dataComparator = y;
this.tree = new LinkedListWithTwoLinks();
}
private int compare(BinarySearchTreeElementInterface s, Object data)
{
return this.dataComparator.compare(s, data);
}
public boolean insert(Object data)
{
boolean success;
BinarySearchTreeElementInterface current;
BinarySearchTreeElementInterface parent;
current = getRoot();
parent = null;
success = false;
if (current == null)
{
getTree().insert(data);
return true;
}
else
{
SearchResult insert;
insert = searchFor(data);
//if (data == "shaker") {System.out.println(insert.resultOfCompare); }
while (current != null)
{
if (insert.insertAsLeftChild())
{
//if (data == "card") {System.out.println("IN RIGHT");}
//System.out.println("IN LEFT");
parent = current;
current = current.getLeftChild();
}
else if (insert.insertAsRightChild())
{
//if (data == "card") {System.out.println("IN RIGHT");}
parent = current;
current = current.getRightChild();
}
}
if (insert.insertAsLeftChild())
{
//parent.setLeftChild(insert.getParentOfLocation()); //insert.getParentOfLocation()
//System.out.println(data);
getTree().insertUsingPrior(parent, data);
//System.out.println(insert.getParentOfLocation()+" bye left");
// System.out.println(insert.getLocation()+" hi");
success = true;
}
else if (insert.insertAsRightChild())
{
//parent.setRightChild(insert.getParentOfLocation());
//System.out.println(data);
getTree().insertUsingNext(parent, data);
//System.out.println(insert.getParentOfLocation()+" bye right");
// System.out.println(insert.getLocation());
success = true;
}
else {success = false;}
/*
figures out if it should be inserted as a left or right child
then call insert using prior/next
}*/
}
return success;
}
private SearchResult searchFor(Object data)
{
/*returns either to node containing the data or the parent of the node of which the data would be a child of*/
if (getTree() == null) {throw new ListEmptyException("Tree is empty!");}
BinarySearchTreeElementInterface currentLocation;
BinarySearchTreeElementInterface parent;
SearchResult destination;
parent = getRoot();
currentLocation = parent;
while (currentLocation != null)
{
if (currentLocation.getData() == data)
{
return new SearchResult(parent, currentLocation, compare(currentLocation, data));
}
if (compare(currentLocation, data) < 0)
{
//System.out.println("IN LEFT");
parent = currentLocation;
currentLocation = currentLocation.getLeftChild();
}
else if (compare(currentLocation, data) > 0)
{
//System.out.println("IN RIGHT");
parent = currentLocation;
currentLocation = currentLocation.getRightChild();
}
}
destination = new SearchResult(parent, currentLocation, compare(parent, data));
//System.out.println(destination.resultOfCompare);
return destination;
/*
* use nothing but BSTEIs
*/
}
public void inOrder()
{
inOrder(getRoot());
}
public void inOrder(BinarySearchTreeElementInterface BSTroot)
{
//System.out.println(BSTroot.getRightChild());
if (BSTroot != null)
{
inOrder(BSTroot.getLeftChild());
System.out.println(BSTroot.getData());
inOrder(BSTroot.getRightChild());
}
/*if (BSTroot.getLeftChild() != null)
{
}
System.out.println(BSTroot.getData());
if (BSTroot.getRightChild() != null)
{
inOrder(BSTroot.getRightChild());
//System.out.println(BSTroot.getData());
}
System.out.println(BSTroot.getData());*/
}
public int size()
{
return tree.size();
}
/*SEARCH RESULT CLASS-----------------------------------------------------------------------------------------*/
public class SearchResult
{
BinarySearchTreeElementInterface location;
BinarySearchTreeElementInterface parentOfLocation;
int resultOfCompare;
public SearchResult(BinarySearchTreeElementInterface parent, BinarySearchTreeElementInterface locate, int comp)
{
this.parentOfLocation = parent;
this.location = locate;
this.resultOfCompare = comp;
}
public BinarySearchTreeElementInterface getLocation()
{
return this.location;
}
public BinarySearchTreeElementInterface getParentOfLocation()
{
return this.parentOfLocation;
}
public boolean insertAsLeftChild()
{
if (resultOfCompare > 0) {return true;}
else {return false;}
}
public boolean insertAsRightChild()
{
if (resultOfCompare < 0) {return true;}
else {return false;}
}
public boolean locationIsLeftOfParent()
{
return this.location == parentOfLocation.getLeftChild();
}
public boolean locationisRightOfParent()
{
return this.location == parentOfLocation.getRightChild();
}
public boolean wasSearchSuccessful()
{
return this.parentOfLocation == this.location;
}
public void setLocation(BinarySearchTreeElementInterface newLocation)
{
this.location = newLocation;
}
public void setLocationOfParent(BinarySearchTreeElementInterface newParentLocation)
{
this.parentOfLocation = newParentLocation;
}
}
}
答案 0 :(得分：0)
BinarySearchTree的compare（x，y）方法是错误的。 它将节点与数据进行比较，并使用比较器将数据与数据进行比较。改变对象＆＃39;到＆＃39; String＆＃39;当您的样本数据是字符串时，它将无法编译。
这应该解决它：
private int compare(BinarySearchTreeElementInterface s, Object data)
{
return this.dataComparator.compare(s.getData(), data);
}
答案 1 :(得分：0)
在你的SearchResult类中，你是否有决定左或右插入交换的方式？
public boolean insertAsLeftChild()
{
if (resultOfCompare > 0) {return true;}
else {return false;}
}
如果比较大于0，它应该对正确的孩子感兴趣，对吗？