algorithm - Worst case in Max-Heapify - How do you get 2n/3?

Question

In CLRS, third Edition, on page 155, it is given that in MAX-HEAPIFY,

The children’s subtrees each have size at most 2n/3—the worst case occurs when the bottom level of the tree is exactly half full.

I understand why it is worst when the bottom level of the tree is exactly half full. And it is also answered in this question worst case in MAX-HEAPIFY: "the worst case occurs when the bottom level of the tree is exactly half full"

My question is how to get 2n/3?

Why if the bottom level is half full, then the size of the child tree is up to 2n/3?

How to calculate that?

Thanks

Answer 1

In a tree where each node has exactly either 0 or 2 children, the number of nodes with 0 children is one more than the number of nodes with 2 children.{Explanation: number of nodes at height h is 2^h, which by the summation formula of a geometric series equals (sum of nodes from height 0 to h-1) + 1; and all the nodes from height 0 to h-1 are the nodes with exactly 2 children}

    ROOT
  L      R
 /     / 
/     /   
-----  -----
*****

Let k be the number of nodes in R. The number of nodes in L is k + (k + 1) = 2k + 1. The total number of nodes is n = 1 + (2k + 1) + k = 3k + 2 (root plus L plus R). The ratio is (2k + 1)/(3k + 2), which is bounded above by 2/3. No constant less than 2/3 works, because the limit as k goes to infinity is 2/3.