Need help on deleting a binary search tree node

Question

This is a grade 12 assignment. One of the questions asks us to write a method in BTree class that takes either a BNode or an integer, and removes the node from the tree.

Here's what I've tried:

public void delete(BNode b){
    if(b==null){
        return;
    }
    if(b.getLeft()==null && b.getRight()==null){

        b = null;
    }
    else{
        //System.out.println("boiboi");
        BNode tmp = b;
        b = null;
        add(tmp.getLeft());
        add(tmp.getRight());
        tmp = null;
    }
}
public void delete(int v){
    //System.out.println("gord");
    delete(find(v));
}

Here's the add and find method which i think are correct:

public BNode find(int v){
    return find(v, root);
}
public BNode find(int v, BNode branch){
    if(branch == null || branch.getVal() == v){
        return branch;
    }
    if(v<branch.getVal()){
        return find(v, branch.getLeft());
    }
    else{//else if(v>branch.getVal())
        return find(v, branch.getRight());
    }
}
public void add(int v){
    if(root == null){
        root = new BNode(v);
    }
    else{
        add(v, root);
    }
}
public void add(int v, BNode branch){
    if(v == branch.getVal()){
        return;
    }
    if(v<branch.getVal()){
        if(branch.getLeft() == null){
            branch.setLeft(new BNode(v));
        }
        else{
            add(v, branch.getLeft());
        }
    }
    else{
        if(branch.getRight() == null){
            branch.setRight(new BNode(v));
        }
        else{
            add(v, branch.getRight());
        }
    }
}
public void add(BNode n){
    if(n==null){
        return;
    }
    add(n.getVal());
}

Here's my testing class:

    BTree bTree = new BTree();
    bTree.add(50);
    bTree.add(60);
    bTree.add(40);
    bTree.add(35);
    bTree.add(55);
    bTree.add(45);
    bTree.add(51);
    bTree.delete(60);
    bTree.display();

the output is still everything i've added: 35 40 45 50 51 55 60 even if i tried to delete 51 which is the simplest case, still same output. Any help or suggestions would be appreciated, thank you.

BNode b is a REFERENCE. b=null won't modify The VALUE of your node, just put a null reference to b. You have modify Parent's left/right reference. If it is root, you have modify root reference. — Chengxi Li
– Chengxi Li, Commented Mar 27, 2013 at 3:52

smk · Accepted Answer · 2013-03-27 03:50:49Z

2

There are three cases that you need to take care of when deleting a node from a BST.

If its a leaf, just go ahead and delete.
If a node has just 1 child , simply connect its child to its parent. [you will obviously need some helper methods to getParentOfNode etc]
If a node has 2 children, find the smallest element in the right subtree.And put its value in the current node and delete that node.

More info here. http://www.cs.sunysb.edu/~skiena/373/notes/lecture6/lecture6.html

answered Mar 27, 2013 at 3:50

smk

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Comments

jaxkewl · Accepted Answer · 2016-08-28 13:09:38Z

/* Algorithm to delete a record from a B-tree of order n.
The field used indicates how many keys in the node are being used. */
    q = NULL;
    p = tree;
    while (p) {
        i = nodesearch(p, key);
        q = p;
        if (i < used(p) -1 && key == k(p,i)) {
            found = TRUE;
            position = i;
            break;
        }
        p = son(p,i);
    }
    if (!found)
    /* error - item not found */
    else if (subtree(p)) {
        /* node is not a leaf */
        if (used(p) > ((n-1)/2)+1)
            /* no underflow */
            delkey (p, position, key);
        else {
            /* Statements to replace key to be deleted */
            /* with successor in father node. */
            replace (p, position, fsucc(p));
            p0 r1 p1 r2 p2 r3 ……. pn-1 rn-1 pn
            q = &fsucc(p);
            qpos = index of fsucc;
            if (used(rbrother(p)) > ((n-1)/2)+1)
                replace (q, qpos, sonsucc(q));
            else
                while (q && used(q) < (n-1)/2) {
                    concatenate(q, brother(q));
                    q = father(q);
                }
        }
    }
    else /* is a leaf */
    delkey(p, position, key);
}

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