Height is $\Omicron(\lg N)$
- Show the height of BST that maintains balance property is $\Omicron(\lg N)$.
This section is an optional reading!
In a BBST, unlike a regular BST, it is guaranteed that the tree's height is $\Omicron(\lg N)$. This claim can be easily observed for a perfect BST, where each internal node has two children and all the leaves are at the same level.
Let's observe the perfect BSTs above
| Height | Nodes | Binary Log Calculation |
|---|---|---|
| $0$ | $1$ | $\log_2 1 = 0$ |
| $1$ | $3$ | $\log_2 3 = 1$ |
| $2$ | $7$ | $\log_2 7 = 2$ |
| $3$ | $15$ | $\log_2 15 = 3$ |
Given these perfect BSTs, one can intuit how this holds more generally.
First, convince yourself that the worst height of a BST is when it has the minimum number of nodes concerning its height. Then, let's call $N(h)$ the minimum number of nodes given height.
The worst case $h \in \Omicron(n)$ is when $N(h) = h+1$.
If we find the upper bound of $N(h)$ for BBST, then we've bounded the height of all BBSTs.
Let's consider a BBST of height $h \geq 3$ and the minimum number of nodes $n=N(h)$. This tree is composed of a root and two subtrees. Since the whole tree has the minimum number of nodes for its height, so do the subtrees. For the big tree to be of height $h$, one of the subtrees must be of height $h-1$. To get the minimum number of nodes, the other subtree is of height $h-2$.
Why can't the other subtree be of height $h-3$ or $h-4$?
Because the balance property requires the height of siblings to differ by at most $1$.
So far, we established for $h \geq 3$,
$$ N(h) = 1 + N(h-1) + N(h-2) $$
We know
$$ N(h) > N (h-1) \space \text{and} \space N(h-1) > N(h-2) $$
therefore
$$ N(h) > N(h-1) + N(h-2) > 2\times N(h-2) $$
If $N(h) > 2\times N(h-2)$ then $N(h-2) > 2\times N(h-4)$, therefore
$$ N(h) > 2\times N(h-2) > 4 \times N(h-4) $$
We can keep doing this
$$ N(h) > \dots > 4\times N(h-4) > 8\times N(h-8) > \dots $$
What does the above expression demonstrate? Why is that significant?
This expression demonstrates that the number of nodes at least doubles each time the height increases by a factor of $2$. This observation is important because $N(h)$ is exponential in $h$. Therefore $h$ is logarithmic in the number of nodes.
Continuing the proof, we see that this can be written as:
$$ N(h) > 2^i \times N(h - 2i) $$
The largest $i$ we can put in there is $i = \frac{h}{2} - 1$:
Why is this value of $i$ chosen as $\frac{h}{2} - 1$?
This allows the expression to be equal to either $1$ or $2$, depending on if $h$ is even or odd.
- Even $h$: $h - 2i = h - (h - 2) = 2$
- Odd $h$: $h - 2i = h - (2(\frac{h}{2}) - 2) = h - ((h + 1)-2) = 1$
So we have:
$$ N(h) > 2^i \times (h-2i) > 2^{\frac{h}{2}} \times N(h - h) > 2^{\frac{h}{2}} $$
So we showed the minimum number of nodes, $n=N(h)$ in a BBST of height $h$, is at least $2^{\frac{h}{2}}$:
$$ n > 2^{\frac{h}{2}} $$
What does this say about height, $h$? First, let's take the logarithm of both sides:
$$ \lg n > \lg 2^{\frac{h}{2}} \Rightarrow h < 2\lg(n) $$
So, $h \in \Omicron(\lg n)$.