问题
I'm working on the case where the node to be deleted is a node. I'm not sure if I need to keep track of the parent so that when I find the node to delete so I can set its parents pointer to null. But then how would I know which child the node to be deleted is? Do I need more if statements?
Any help is appreciated, I feel its not too complicated but I'm just confused on how to actually get rid of the node.
This is what I have so far:
public void insert(E s)
{
root = insert(s, root);
}
private Node<E> insert(E s, Node<E> T)
{
//easiest case, empty tree, create new tree
if(T == null)
{
T = new Node<E>(s);
}
//easiest case, found s
else if(s.compareTo(T.getData()) == 0)
{
System.out.println("Item already present.");
}
//s is greater than T, insert on right subtree
else if(s.compareTo(T.getData()) > 0)
{
T.setRight(insert(s, T.getRight()));
}
//s is less than T, insert on left subtree
else
{
T.setLeft(insert(s,T.getLeft()));
}
return T;
}
public void delete(E d)
{
delete( d, root);
}
private void delete( E d, Node<E> T)
{
if(T == null)
{
}
else if(d.equals(T.getData()))
{
System.out.println("it found the node at least");
if(T.getRight() == null && T.getLeft() == null)
{
}
//code other cases for a node with one child and node with two children
}
else if(d.compareTo(T.getData()) > 0)
{
System.out.println("going right");
delete(d, T.getRight());
}
//s is less than T, insert on left subtree
else
{System.out.println("going left");
delete(d,T.getLeft());
}
}
回答1:
public Node<E> search(Node<E> node, E d)
{
while(node!=null && ((node.getLeft()!=null && !node.getLeft().getData().equals(d)) || (node.getRight()!=null && !node.getRight().getData().equals(d)))
{
if(d.compareTo(node.getData()) < 0)
{
node = node.getLeft();
}
else
{
node = node.getRight();
}
}
return node;
}
private void delete(E d)
{
// Search the node
Node parent = search(root, d);
// parent is the parent node under which the required node is present
// Now check which child it is - left or right
if(parent == null)
{
System.out.println("Element not found");
return;
}
if(parent.getLeft().getData().equals(d))
{
// Left child
parent.setLeft(null);
}
else
{
// Right child
parent.setRight(null);
}
}
NOTE: If you would had parent node in the Node<E> then it would have been very easy.
回答2:
Track the parent node as you traverse the tree like this if you don't want to change the Node structure.
private Node<E> parent;
public void insert(E s)
{
root = insert(s, root);
}
private Node<E> insert(E s, Node<E> T)
{
//easiest case, empty tree, create new tree
if(T == null)
{
T = new Node<E>(s);
}
//easiest case, found s
else if(s.compareTo(T.getData()) == 0)
{
System.out.println("Item already present.");
}
//s is greater than T, insert on right subtree
else if(s.compareTo(T.getData()) > 0)
{
T.setRight(insert(s, T.getRight()));
}
//s is less than T, insert on left subtree
else
{
T.setLeft(insert(s,T.getLeft()));
}
return T;
}
public void delete(E d)
{
delete( d, root);
}
private void delete( E d, Node<E> T)
{
if(T == null)
{
}
else if(d.equals(T.getData()))
{
System.out.println("it found the node at least");
if(T.getRight() == null && T.getLeft() == null)
{
if (parent != null){//For the first node, parent will be null
if (d.equals(parent.getRight().getData())){//Data matches with the right node of parent
parent.setRight(null);
}else{//Data matches with the left node of parent
parent.setLeft(null);
}
//Reset parent node
parent = null;
}
}
//code other cases for a node with one child and node with two children
}
else if(d.compareTo(T.getData()) > 0)
{
parent = T;// Make the current node as parent
System.out.println("going right");
delete(d, T.getRight());
}
//s is less than T, insert on left subtree
else
{
parent = T;// Make the current node as parent
System.out.println("going left");
delete(d,T.getLeft());
}
}
来源：https://stackoverflow.com/questions/15170999/java-binary-search-tree-delete-node-with-no-children