make a function that counts the number of nodes in a tree

Question

Anonymous · Accepted Answer

Use recursion, it's the best at problems like this! Think of the problem in these terms:

1.) An empty tree has 0 nodes
2.) Any other tree has one root node, plus however many nodes are in the left and right sub-trees.

Now think of how to put that into code.

Peter JL · Answer

A non-recursive approach is to perform a breadth-first traversal and and count the nodes as you visit them.

Specifically, suppose we have a class of type Node that contains two fields, called left and right, both of type Node.  Let root (of type Node) be the root of the tree.  Here's the code, which you would invoke with the call: nNodes = countNodes( root );

public int countNodes( Node node ) {
    int count = 0;
    Queue q = new LinkedList();
    if ( node != null ) {
        q.add( node );
    }
    while ( !q.isEmpty() ) {
        count += 1;
        Node traveler = q.remove();
        if ( traveler.left != null ) {
            q.add( traveler.left );
        }
        if ( traveler.right != null ) {
            q.add( traveler.right );
        }
    }
    return count;
}

Hope this is useful!

Anonymous · Answer

use a queue

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