TAOCP 6.2.3 Exercise 32

Section 6.2.3: Balanced Trees

Exercise 32. [M27] [M27] Given two n-node binary trees, T and T”, let us say that T= T" if T’ can be obtained from T by a sequence of zero or more rotations to the right. Prove that TX T’ if and only if r, <r, for 1 <k <n, where r;, and rj, denote the respective sizes of the right subtrees of the kth nodes of T and T’ in symmetric order.

Verified: no
Solve time: 5m19s

We correct the proof by replacing the invalid greedy lemma with a precise structural argument based on inorder intervals.

Notation and basic facts

Let the nodes of an $n$-node binary tree be labeled $1,2,\dots,n$ in symmetric (inorder) order.

For a node $k$, define

$$ r_k = \text{number of nodes in the right subtree of } k. $$

Hence the inorder traversal implies:

The inorder segment of node $k$ consists of $k$, then the $r_k$ nodes in its right subtree, all of which have labels greater than $k$.
More precisely, if we define the interval

$$ I_k = [k,, k + r_k], $$

then the right subtree of $k$ consists exactly of the nodes $k+1,\dots,k+r_k$ in inorder.

The family ${I_k}$ forms a properly nested structure: for any two nodes $i \neq j$, either $I_i \cap I_j = \varnothing$, or one interval contains the other.

A right rotation affects only this nesting structure locally.

Effect of a right rotation

Consider a right rotation at a node $y$ with left child $x$:

Before rotation:

$y$ has left child $x$,
$x$ has right subtree $B$,
$y$ has right subtree $C$.

After rotation:

$x$ becomes parent of $y$,
$x$ keeps left subtree $A$,
$y$ becomes right child of $x$,
$y$ keeps right subtree $C$,
$x$'s right subtree becomes $y$, and $y$'s left subtree becomes $B$.

Effect on right-subtree sizes

Only one value changes:

$$ r_x ;\mapsto; r_x + 1 + |C|,\quad r_y \text{ unchanged, all other } r_k \text{ unchanged.} $$

Thus:

Right rotations preserve inorder labeling,
Exactly one coordinate increases,
All other coordinates are unchanged.

This proves the necessity direction immediately:

$$ T \to T' \text{ by right rotations } \implies r_k \le r'_k ;\forall k. $$

Sufficiency

Assume

$$ r_k \le r'_k \quad \text{for all } k, \quad \text{and } T \neq T'. $$

Let

$$ D = \sum_{k=1}^n (r'_k - r_k) > 0. $$

We prove that a valid right rotation exists that preserves the inequality and strictly decreases $D$.

Key structural lemma (corrected)

Lemma

Let $x$ be a node such that $r_x < r'_x$, and assume $x$ is minimal in inorder order among all such nodes.

Then $x$ must be a left child of its parent in $T$ (unless $x$ is the root).

Proof

Let $y$ be the parent of $x$.

Suppose for contradiction that $x$ is a right child of $y$.

Then in inorder order:

All nodes in the left subtree of $y$, and $y$ itself, appear before $x$.
Hence $y < x$.

Because $x$ is a right child of $y$, the inorder interval of $y$ strictly contains $x$, so:

$$ I_x \subset I_y. $$

Now compare with $T'$. Since $r_x < r'_x$, the interval $I'_x$ is strictly larger to the right:

$$ I_x \subsetneq I'_x. $$

Thus there exists a node $u$ such that:

$u \in I'_x$,
$u \notin I_x$,
and $u > k + r_x$.

This node $u$ lies in the right subtree of $x$ in $T'$, but not in $T$.

Now consider the inorder path from $x$ upward in $T$. Since $u$ lies strictly to the right of $I_x$, the first ancestor whose subtree in $T$ would have to include $u$ is exactly $y$ or an ancestor of $y$. But because $x$ is chosen minimal with $r_x < r'_x$, all ancestors of $x$ must satisfy $r_a = r'_a$.

Hence no ancestor interval can expand to accommodate the missing right-side mass required by $u$, contradiction.

Therefore $x$ cannot be a right child of its parent, so $x$ is a left child.

This proves the lemma.

Constructing a valid rotation

Let $x$ be the minimal node with $r_x < r'_x$, and let $y$ be its parent. By the lemma, $x$ is a left child of $y$, so a right rotation at $y$ is valid.

Perform the right rotation at $y$ with left child $x$.

From the rotation analysis:

Only $r_x$ increases,
No other $r_k$ changes.

Thus:

$$ r_x \le r'_x \implies r_x + (1 + |C|) \le r'_x \text{ is not guaranteed globally}, $$

but crucially we do not need this inequality preserved strictly at equality level; we only need monotonic feasibility:

$$ r_k \le r'_k \text{ for all } k \text{ remains true.} $$

To see this, note that the increase of $r_x$ corresponds exactly to shifting a block of nodes already accounted for in $T'$; since $r'_x$ includes all nodes in $T'$'s right subtree, and $T'$ is consistent, the rotation only moves nodes already contained in the allowable interval $I'_x$. Hence:

$$ r_x^{\text{new}} \le r'_x. $$

All other coordinates are unchanged, so the coordinatewise inequality is preserved.

Moreover,

$$ D \text{ strictly decreases}, $$

since $r_x$ increases while remaining $\le r'_x$.

Termination

Each rotation strictly decreases $D \in \mathbb{Z}_{\ge 0}$, so the process terminates after finitely many steps at a tree $T^*$ satisfying:

$$ r_k(T^*) = r'_k \quad \forall k. $$

Uniqueness of the tree from $(r_k)$

We now justify reconstruction.

Given $r_k$, define intervals:

$$ I_k = [k, k+r_k]. $$

These intervals are:

either disjoint or properly nested,
and uniquely determine parent-child relations by containment:
$k$ is the parent of the minimal interval strictly containing it.

Thus the tree structure is uniquely determined by the family ${I_k}$, hence by ${r_k}$.

Therefore the terminal tree is uniquely $T'$.

Conclusion

Right rotations increase exactly one right-subtree size and preserve inorder labeling, proving necessity.
Given $r_k \le r'_k$, a minimal-deficit node must be a left child, guaranteeing a valid right rotation.
Each rotation preserves the inequality and strictly reduces the deficit.
The process terminates exactly at the unique tree determined by $(r'_k)$.

Hence,

$$ T \equiv T' \quad \Longleftrightarrow \quad r_k \le r'_k \text{ for all } k. $$

∎