how to find the position of right most node in last level of complete binary tree?

Question

I am doing a problem in binary trees, and when I came across a problem find the right most node in the last level of a complete binary tree and the issue here is we have to do it in O(n) time which was a stopping point, Doing it in O(n) is simple by traversing all the elements, but is there a way to do this in any complexity less than O(n), I have browsed through internet a lot, and I couldn't get anything regarding the thing. Thanks in advance.

Answer 1

I assume that you know the number of nodes. Let n such number.

In a complete binary tree, a level i has twice the number of nodes than the level i - 1.

So, you could iteratively divide n between 2. If there remainder then n is a right child; otherwise, is a left child. You store into a sequence, preferably a stack, whether there is remainder or not.

Some such as:

Stack<char> s;
while (n > 1)
{
  if (n % 2 == 0)
    s.push('L');
  else
    s.push('R');
  n = n/2; // n would int so division is floor
}

When the while finishes, the stack contains the path to the rightmost node.

The number of times that the while is executed is log_2(n).

Answer 2

Yes, you can do it in O(log(n)^2) by doing a variation of binary search.

This can be done by first going to the leftest element¹, then to the 2nd leftest element, then to the 4th leftest element, 8th ,... until you find there is no such element. Let's say the last element you found was the ith, and the first you didn't was 2i.

Now you can simply do a binary search over that range.

This is O(log(n/2)) = O(logn) total iterations, and since each iteration is going down the entire tree, it's total of O(log(n)^2) time.

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(1) In here and the followings, the "x leftest element" is referring only to the nodes in the deepest level of the tree.