Intersection of 2 binary search trees

Question

Hey, So I want to create a new tree which is basically the intersection (mathematical definition of intersection) of 2 given binary search trees. I have a method that prints out all the nodes at a particular level of the tree and I have a method that can find out the depth of the tree.I am pasting my work so far though it is incomplete and I'm stuck with the logic.Help will be appreciated.

    public static Bst<Customer> intersect (Bst<Customer> a, Bst<Customer> b){
    Bst<Customer> result = new Bst<Customer>();
    BTNode<Customer> cur1;
    BTNode<Customer> cur2;
    BTNode<Customer> cur3;
    cur1=a.root;
    cur2=b.root;
    cur3=result.root;
    int Resultdepth;
    if(a.maxDepth()<b.maxDepth())
        Resultdepth=a.maxDepth();
    else
        Resultdepth=b.maxDepth();

    if(cur1==null || cur2==null){ // Handeling the root case intially
        result = null;
    }
    else 
      cur3.item.set_account_id(cur1.item.get_accountid()+ cur2.item.get_accountid());

    cur1=cur1.left;
    cur2=cur2.left;
    cur3=cur3.left;       

    while(<some check>){

    }


    return result;

}


    public int maxDepth(){
        return mD(root);
    }

    int mD(BTNode<E> node){
       if (node==null) {
            return(0);
        }
       else {
            int lDepth = mD(node.left);
            int rDepth = mD(node.right);
            // use the larger + 1
            return(Math.max(lDepth, rDepth) + 1);
        }
    }

     // for printing the nodes at a particular level and giving the starting level
      public void PrintAt(BTNode<E> cur, int level, int desiredLevel) {
         if (cur == null) {
            return;
        }
         if (level == desiredLevel) {
             System.out.print(cur.item.toString() + "");
          }
         else {
             PrintAt(cur.left, level+1, desiredLevel);
             PrintAt(cur.right, level+1, desiredLevel);
          }
}

I don't see anything related to "intersection" in your code... — Hosam Aly, Apr 20 '11 at 10:30
I strongly advise that you don't try to hard-code any max depth by using discrete variables like `cur1, cur2, cur3` etc. Your solution must work for *all* cases. You'll find that the solution to the general case is also simpler than one that makes assumptions. — BoffinBrain, Apr 20 '11 at 15:10

score 3 · Accepted Answer · answered Apr 20 '11 at 11:19

3

You have to traversal both trees in order at the same time and "in sync".

I'd suggest to implement the Iterable interface for your class. Then you get the first values from both trees. If they are equal, put it in the new tree, and get the next values from both iterators. If not, iterate the iterator with the smaller values until the value you get is as least as big as the last value from the other iterator. Rinse and repeat.

answered Apr 20 '11 at 11:19

Landei

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That's what I would have done as well. I can't think of a more efficient solution (except, perhaps if you knew more about the average density of the data, you could combine basic tree traversal with partial binary searches when you come across a large enough gap). – BoffinBrain Apr 20 '11 at 15:08
uhmm i don't really know how to implement the iterable class but i coded something and currently debugging it. I posted a new question here – dawnoflife Apr 20 '11 at 19:51

score 1 · Answer 2 · answered Sep 17 '12 at 13:58

For the recursive implementation of finding intersection of two binary search trees , I came up with the following code. I am not very sure of the time complexity, but it does work all right.

void BST::findIntersection(cell *root1, cell * root2) {

if(root1 == NULL ) { 
//  cout<<"Tree 1 node is null , returning"<<endl;  
    return;
}
if(root2 == NULL) {
//  cout<<"Tree 2 node is null , returning"<<endl;  
    return;
}
//cout<<"Comparing tree 1 : "<<root1->data<< "   and tree 2 : " << root2->data<<endl;
if(root1->data==root2->data) {
//  cout<<"tree 1 equal to tree 2 "<<endl;
    insert(root1->data);
//  cout<<"Inserting in new tree : "<<root1->data<<endl;
    findIntersection(root1->left,root2->left);
    findIntersection(root1->right, root2->right);
}
else if(root1->data>root2->data) {
//  cout<<"tree 1 > tree 2 "<<endl;
    findIntersection(root1,root2->right);
    findIntersection(root1->left, root2);
}
else  {
//  cout<<"tree 1 < tree 2 "<<endl;
    findIntersection(root1->right,root2);
    findIntersection(root1, root2->left);
}

}

score 0 · Answer 3 · answered Apr 20 '11 at 10:33

0

The intersection of two trees is presumably the nodes that are in both trees?

Given that you'll have to explore the tree to do this, why not just do an in-order traversal, store the nodes and then do an intersection operation on ordered lists?

answered Apr 20 '11 at 10:33

Jeff Foster

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Jeff, Yes, it is intersection is the nodes in the 2 trees. I don't really want to use lists in this case. I know this can be done through conditional checks and am looking to be able to do that. – dawnoflife Apr 20 '11 at 10:41

Michael · Answer 4 · 2011-04-20T21:45:29.640

0

My suggestion for such an intersection is simple:

Given tree A and tree B, to find tree C = A \intersect B:

1: Copy either tree A or B. Let us assume A for clarity.
This copy is now your tree C. Now let's 'trim' it.
2: For c = C.root_node and b = B.root_node:
if b==c,
Repeat the procedure with nodes b.left, c.left
Repeat the procedure with nodes b.right, c.right
else,
Remove c (thereby removing all subsequent children, it is implied they are unequal)

If this implementation would work, it would avoid the use of iterators and the like, and boil down to a simple recursive traversal. (Like this!)

Ask if you would like further clarification.

Regards.

edited Apr 20 '11 at 21:45

answered Apr 20 '11 at 21:38

Michael

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why do you do "if b==c". i don't see what you are trying to achieve with that – dawnoflife Apr 21 '11 at 03:00
I took an opposite approach. Copy EVERYTHING then delete difference. – Michael Apr 23 '11 at 22:52
@Michael, Let's say that there are two trees A and B. A.root->data = 5, A.root->left->data = 6, A.root->right->data = 8 B.root->data = 5, B.root->left->data = 7, B.root->right->data = 8 According to your approach there would be no intersection – Anubhav Agarwal Nov 03 '12 at 18:46

Intersection of 2 binary search trees

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