我正在尝试为二叉搜索树编写代码,我正在处理的第一个方法是add(insert)方法。 root似乎正确插入,但是在添加第二个节点时我得到空指针异常。我将在我的代码中用注释指出确切的问题点。
如果你能看到如何解决这些问题,或者让我知道我的整体逻辑是否存在缺陷,那将会非常有帮助.--我会提到这是针对学校的,所以我不打算做一个真的令人印象深刻的模型...我的大多数布局选择只是反映了我们在课堂上工作的方式。此外,方法名称由教师选择,应保持不变。随意编辑格式,有点麻烦。
public class BinarySearchTree
{
private static Node root;
public BinarySearchTree()
{
root = null;
}
public static void Add (Node newNode)
{
Node k = root;
if (root == null)//-----------------IF TREE IS EMPTY -----------------
{
root = newNode;
}
else // -------TREE IS NOT EMPTY --------
{
if (newNode.value > k.value) //-------NEW NODE IS LARGER THAN ROOT---------
{
boolean searching = true;
while(searching) // SEARCH UNTIL K HAS A LARGER VALUE
{ //***CODE FAILS HERE****
if(k.value > newNode.value || k == null)
{
searching = false;
}
else {k = k.rightChild; }
}
if ( k == null) { k = newNode;}
else if (k.leftChild == null){ k.leftChild = newNode;}
else
{
Node temp = k.leftChild;
k.leftChild = newNode;
newNode = k.leftChild;
if(temp.value > newNode.value )
{
newNode.rightChild = temp;
}
else
{
newNode.leftChild = temp;
}
}
}
if (newNode.value < k.value) //-----IF NEW NODE IS SMALLER THAN ROOT---
{
boolean searching = true;
while(searching) // ----SEARCH UNTIL K HAS SMALLER VALUE
{// **** CODE WILL PROBABLY FAIL HERE TOO ***
if(k.value < newNode.value || k == null) {searching = false;}
else {k = k.leftChild;}
}
if ( k == null) { k = newNode;}
else if (k.rightChild == null){ k.rightChild = newNode;}
else
{
Node temp = k.rightChild;
k.rightChild = newNode;
newNode = k.rightChild;
if(temp.value > newNode.value )
{
newNode.rightChild = temp;
}
else
{
newNode.leftChild = temp;
}
}
}
}} // sorry having formatting issues
}
public class Node
{
int value;
Node leftChild;
Node rightChild;
public Node (int VALUE)
{
value = VALUE;
}
}
public class TestIT
{
public static void main(String[] args)
{
BinarySearchTree tree1 = new BinarySearchTree();
Node five = new Node(5);
Node six = new Node(6);
tree1.Add(five);
tree1.Add(six);
System.out.println("five value: " + five.value);
System.out.println("five right: " + five.rightChild.value);
}
}
答案 0 :(得分:1)
从左到右检查条件语句,因此在检查k.value&gt;之前,您需要检查k是否为null。 newNode.value因为如果k为null,则它没有值。