Question

I want to calculate the tighter bound on heapify using siftdown approach so I proceeded as follows :

At each level i, each key on that level can go to leaf level h (where h is height of tree) in the worst case.
As there are 2ⁱ nodes at level i, so total work done is

∑_0≤i≤h (h – i ) * 2ⁱ

But I couldn’t proceed further. I know it has to come O(n) but I couldn’t reach it. Please help me solving this.

Answer 1

S = ∑_0≤i≤h (h – i ) * 2ⁱ

S = h + 2(h – 1) + 4(h – 2) + … + 2^{h – 1} … (1)

Multiply both sides by 2,

2S = 2h + 4(h – 1) + 8(h – 2) + … + 2^h … (2)

Subtract from (1) from (2),

S = h -2h + 2 + 4 + 8 + …. + 2^h

= -h – 1 + (1 + 2 + 4 + 8 +… + 2^h )

= (2^{h + 1} – 1) – (h + 1)

[Note: 1 + 2 + 4 + 8 + … + 2^h = (2^{h + 1} – 1)

Since a complete binary tree of height h has between 2h and 2h + 1 nodes, the above sum is O(n) where n is the number of nodes in the heap.