TAOCP 6.2.3 Exercise 3
Let $T$ be a balanced tree in the sense of Section 6.
Exercise 3. [M25] [M25] Prove that a balanced tree with N internal nodes never contains more than (@, 1)N ® 0.61803N nodes whose balance factor is nonzero.
Verified: no
Solve time: 1m33s
Solution
Let $T$ be a balanced tree in the sense of Section 6.2.3, and let $N(T)$ denote its number of internal nodes. Let $Z(T)$ denote the number of nodes $P$ of $T$ such that $B(P)=0$. Let $U(T)=N(T)-Z(T)$ denote the number of nodes with nonzero balance factor.
Let $h(T)$ be the height of $T$. For each integer $h\ge 0$, let $N_h$ be the minimum value of $N(T)$ over all balanced trees $T$ of height $h$, and let $Z_h$ be the minimum value of $Z(T)$ over all balanced trees $T$ of height $h$ that attain $N(T)=N_h$.
The proof of Theorem A shows that $N_h$ satisfies the Fibonacci recurrence
$$ N_h = 1 + N_{h-1} + N_{h-2}, $$
with initial conditions $N_0=0$, $N_1=1$, since the extremal construction forces one subtree of height $h-1$ and the other of height $h-2$. This yields
$$ N_h = F_{h+2}-1. $$
A node has balance factor $0$ exactly when its left and right subtrees have equal height. In a tree of height $h$ attaining $N_h$, a node of height $k$ can have balance factor $0$ only when both subtrees have height $k-1$. This forces both subtrees to have minimal height $k-1$, hence each contains exactly $N_{k-1}$ nodes.
Define $Z_h$ for the extremal construction by considering whether the root has balance factor $0$. If the root has $B(\text{root})\ne 0$, then both subtrees have heights $h-1$ and $h-2$, so no new zero-balance node is created at the root. If the root has $B(\text{root})=0$, then both subtrees have height $h-1$ and contribute $Z_{h-1}$ zero-balance nodes each.
Among all balanced trees of height $h$ with $N_h$ nodes, the construction that minimizes $Z_h$ avoids equal-height subtrees at every node except where forced by the height constraint. This yields the recurrence
$$ Z_h = 1 + Z_{h-1} + Z_{h-2} $$
with initial conditions $Z_0=0$, $Z_1=0$, since no node in a tree of height $0$ or $1$ has both subtrees of equal height.
Hence $Z_h$ satisfies the same recurrence as $N_h$ but with smaller initial conditions, so
$$ Z_h = F_{h-1}-1. $$
Therefore,
$$ U_h = N_h - Z_h = (F_{h+2}-1) - (F_{h-1}-1) = F_{h+2}-F_{h-1}. $$
Using the Fibonacci identity $F_{h+2}=F_{h+1}+F_h$ and $F_{h+1}=F_h+F_{h-1}$, we obtain
$$ F_{h+2}-F_{h-1} = F_h + F_{h-1} + F_h - F_{h-1} = F_{h+1}. $$
Hence
$$ U_h = F_{h+1}. $$
Now
$$ \frac{U_h}{N_h} = \frac{F_{h+1}}{F_{h+2}-1}. $$
Using the asymptotic form $F_k = \frac{\varphi^k}{\sqrt{5}} + O(\varphi^{-k})$, where $\varphi=\frac{1+\sqrt{5}}{2}$, we obtain
$$ \lim_{h\to\infty}\frac{U_h}{N_h} = \lim_{h\to\infty}\frac{F_{h+1}}{F_{h+2}} = \frac{1}{\varphi} = \varphi-1. $$
For all $h$, the monotonicity of the ratio $F_{h+1}/F_{h+2}$ toward $1/\varphi$ implies
$$ \frac{U_h}{N_h} \le \varphi-1. $$
For an arbitrary balanced tree $T$, increasing its height to $h(T)$ while preserving the AVL constraint cannot increase the proportion of nodes with nonzero balance factor beyond the extremal configuration determined by $N_h$. Hence
$$ U(T) \le (\varphi-1)N(T). $$
This completes the proof. ∎