Heapify: A Linear-Time Operation

Each node, in Floyd's algorithm, in the worst case, sinks to become a leaf. The sink operation involves repeatedly (recursively) swapping the value with one of the children. Let's make the following observation:

So if you are a leaf, at height $h=0$, you make no swaps down. If you are one level higher, at height $h=1$, you make at most one swaps down. And so on. Let's make the following observation:

In general, there are at most $\left \lceil \frac{n}{2^{h+1}} \right \rceil$ nodes of height $h$, so the cost of building a heap is

$$\sum_{h=0}^{\left \lfloor \lg n \right \rfloor} \left \lceil \frac{n}{2^{h+1}} \right \rceil \Omicron(h)$$

We can rewrite the above expression as

$$O \left ( \sum_{h=0}^{\left \lfloor \lg n \right \rfloor} \left \lceil \frac{n}{2^{h+1}} \right \rceil h \right )$$

Since $n$ does not depend on the value of $h$, it can be moved out of the fraction

$$O \left ( n \sum_{h=0}^{\left \lfloor \lg n \right \rfloor} \left \lceil \frac{h}{2^{h+1}} \right \rceil \right )$$

Further observe that $\left \lceil \frac{h}{2^{h}\times 2} \right \rceil < \frac{h}{2^{h}}$ and we can use the upper-bound within big-O notation.

$$O \left ( n \sum_{h=0}^{\left \lfloor \lg n \right \rfloor} \frac{h}{2^{h}} \right )$$

Moreover,

$$\sum_{h=0}^{\left \lfloor \lg n \right \rfloor} \frac{h}{2^{h}} < \sum_{h=0}^{\infty} \frac{h}{2^{h}}$$

And, we can use the upper-bound within big-O notation: ​​ $$O \left ( n \sum_{h=0}^{\infty} \frac{h}{2^{h}} \right )$$ ​ It turns out the summation converges to $2$: (It is beyond the scope of this course to prove this, but you can check it out on Wolfram Alpha.)

$$\sum_{h=0}^{\infty} \frac{h}{2^{h}} = 2$$

Therefore, we get the following which means the heapify operation is linear time.

$$O \left ( n \times 2 \right ) = O \left ( n \right )$$

A more intuitive analysis

Let's consider a perfect binary tree; here is a less accurate but more intuitive analysis:

Height	#nodes	worst-case #swaps	Total work
$0$	$n/2$	$n/2 \times 0 = 0$	$0$
$1$	$n/4$	$n/4 \times 1 = n/4$	$n/4$
$2$	$n/8$	$n/8 \times 2 = n/4$	$n/2$
$3$	$n/16$	$n/16 \times 3 < n/8$	$n/2 + n/8$
$4$	$n/32$	$n/32 \times 4 = n/8$	$n/2 + n/4$
$5$	$n/64$	$n/64 \times 5 < n/16$	$n/2 + n/4 + n/16$
$6$	$n/128$	$n/128 \times 6 < n/16$	$n/2 + n/4 + n/8$
$\vdots$	$\dots$	$\dots$	$n/2 + n/4 + n/8 + \dots$

The summation for total work is in $\Omicron(n)$.