algorithm - how to determine if the kth largest element of the heap is greater than x

Question

Consider a binary heap containing n numbers (the root stores the greatest number). You are given a positive integer k < n and a number x. You have to determine whether the kth largest element of the heap is greater than x or not. Your algorithm must take O(k) time. You may use O(k) extra storage

Answer 1

Simple dfs can do the job. We have a counter set to zero. Start from the root and in each iteration check the value of current node; if it is greater than x, then increase the counter and continue the algorithm for one of the child nodes. The algorithm terminates if counter is bigger than equal k or if there is no node left to check. The running time is O(k) because at most k node will be iterated and each iteration is in O(1).

A pseudo-code looks like as follows.

    void CheckNode(Node node,int k, int x, ref int counter)
    {
        if (node.value > x)
        {
            counter++;
            if (counter >= k)
                return;

            CheckNode(node.Left, k, x, ref counter);
            CheckNode(node.Right,k, x, ref counter);
        }
    }

use it:

        counter = 0;
        CheckNode(root,index,val,counter );
        if (counter >= index)
            return true;
        return false;

if node.value < x then all children values are smaller than x and there is no need to check.

As @Eric Mickelsen mentioned in comments worst case running time is exactly 2k-1 (k>0) as follows.

The number of nodes visited with values greater than x will be at most k. Each node visited with value less than x must be a child of a visited node with value greater than x. However, because every node visited except the root must have a parent with value greater than x, the number of nodes of value less than x visited must be at most ((k-1)*2)-(k-1) = k-1, since k-1 of the (k-1)*2 children have values greater than x. This means that we visit k nodes greater than x and k-1 less than x, which is 2k-1.