以 O(log n) 获取二叉树的大小

Question

Full disclosure: This is Homework .

I'm asked to return the rank in a Binary Tree, and after some coding I got it to work. But since my code doesn't get accepted, I noticed that the code should run in O(log n)

And the culprit that slowes this down is my size method:

public int size(Node node){
    if (node != null) return (size(node.left) + 1 + size(node.right));
    return 0;
}

Which I call to get the rank of all elements smaller than the one I need to look for.

Now, I googled and all, but apparentely it is not possible to get the size of a BT in log n time?

How would I need to do this then?

Answer 1

With a simple binary tree that only stores children, it isn't possible to get the size in O(log(n)) time since there are n nodes, and you must count every one of them. However, why restrict yourself to only having children? Like many data structures, you can store the size with the node:

class Node {
    public Node left;
    public Node right;
    public int value;
    private int size; // initialize this to 1
}

Then, when inserting into the node, increment the size of every node you encounter:

public void insert(int value) {
    // increment the size
    this.size++;
    if (value < this.value) {
        // obviously check for null as well and insert as appropriate
        this.left.insert(value);
    } else {
        this.right.insert(value);
    }
}

Now, you can get the size in O(1) time, since every node has it.