从二叉搜索树中删除节点

Question

My Binary Search Tree program doesn't seem to be deleting anything when I call the deleteNode method. The BST is built perfectly, its just deleting the node part that doesn't work. I call it from my main like this:

System.out.println("Please enter a number you would like to delete from the tree");
    temp = reader.nextLine();
    try {
        int numTemp = Integer.parseInt(temp);
        TreeNode treeTemp = bst.deleteNode(numTemp, bst.getRoot());
        bst.setRoot(treeTemp);
    }
    catch(Throwable e){
        System.err.println(e);
    }
    bst.printInBST(bst.getRoot());

In my BinarySearchTree class I implement my deleteNode methods as following:

public TreeNode deleteNode(int x, TreeNode temp){
    if(temp != null){
        if(x > (int)((Integer)temp.getValue())){
            temp.setLeft(deleteNode(new Integer(x), temp.getLeft()));
        }
        else if(x < (int)((Integer)temp.getValue())){
            temp.setRight(deleteNode(new Integer(x), temp.getRight()));
        }
        else if(temp.getLeft() != null & temp.getRight() != null){
            TreeNode temp2 = new TreeNode(temp.getRight().getValue());
            while(temp2.getLeft() != null){
                temp2 = temp2.getLeft();
            }
            temp = temp2;
            temp.setRight(remove(temp.getRight()));
        }
    }
    return temp;
}
public TreeNode remove(TreeNode temp){
        if(temp.getLeft() != null){
            temp.setLeft(remove(temp.getLeft()));
            return temp;
        }
        else {
            return temp.getRight();
        }
}

Answer 1

I think you are not handling the

case 1: where the deleting node is a leaf node

case 2: where the deleting node has just 1 child

---

the else if part should be something like this.

else if( temp.getLeft() != null && temp.getRight() != null ) // Two children
{
      temp.setValue( findMin( temp.getRight() ).getValue());
      temp.setRight ( deleteNode( temp.getValue(), temp.getRight() );
}
else
     temp = ( temp.getLeft() != null ) ? temp.getLeft() : temp.getRight();

return temp;

---

The findMin method is to find the inorder successor of the node to be deleted.

private TreeNode findMin( TreeNode t )
{
        if( t == null )
            return null;
        else if( t.getLeft() == null )
            return t;
        return findMin( t.getLeft() );
}

---

I hope this answer your question.

Answer 2

Writing legible code makes bugs easier to spot - both by yourself and others. A first step is choosing more expressive variable names than temp , temp2 , and treeTemp .

Also, it is really not neccessary to do new Integer(x) to assign a method parameter of type int . Simply writing x instead has the same effect, is faster at runtime, and makes it easier to spot the code that matters.

As for bugs, the first one I see is:

TreeNode temp2 = new TreeNode(temp.getRight().getValue());

That creates a copy of the TreeNode. Changing that copy won't affect the original node. Also, the copy probably doesn't have left or right set, since you only pass the value to the constructor. I wonder why think you need a copy? After all, you don't create one here either:

deleteNode(new Integer(x), temp.getRight())

Next, as Sashwat points out, if the node to delete has less than 2 children, your code doesn't do anything, as none of the conditions in deleteNode matches.

Answer 3

  public class BSTNode {
  …

  public boolean remove(int value, BSTNode parent) {
        if (value < this.value) {
              if (left != null)
                    return left.remove(value, this);
              else
                    return false;
        } else if (value > this.value) {
              if (right != null)
                    return right.remove(value, this);
              else
                    return false;
        } else {
              if (left != null && right != null) {
                    this.value = right.minValue();
                    right.remove(this.value, this);
              } else if (parent.left == this) {
                    parent.left = (left != null) ? left : right;
              } else if (parent.right == this) {
                    parent.right = (left != null) ? left : right;
              }
              return true;
        }
  }

Answer 4

Not 100% sure if this is your only problem, but should:

else if(temp.getLeft() != null & temp.getRight() != null)

actually be:

else if(temp.getLeft() != null && temp.getRight() != null)

ie you only have one & for the "and" operation when you should have two?