TAOCP 6.2.2 Exercise 13

Section 6.2.2: Binary Tree Searching

Exercise 13. [M28] [M28] If the root of a random tree is deleted by Algorithm D, is the resulting tree still random?

Verified: yes
Solve time: 4m26s

We restart from the permutation model, since this avoids any circular use of symmetry or exchangeability.

Correct solution

Let $S = {1,\dots,n+1}$. Let $\pi$ be a uniformly random permutation of $S$, and let $T(\pi)$ be the binary search tree produced by Algorithm T (standard BST insertion in permutation order). This is the definition of a random tree in Section 6.2.2.

Let $a = \pi_1$, so $a$ is the root of $T(\pi)$.

Write

$$ L = {x \in S : x < a}, \quad R = {x \in S : x > a}. $$

Then the left and right subtrees of the root are independent random BSTs on $L$ and $R$, respectively.

Let $T'$ be the tree obtained from $T(\pi)$ by deleting the root using Algorithm D. We must show:

$T'$ has the same distribution as $T(\pi')$, where $\pi'$ is a uniformly random permutation of $S \setminus {a}$.

This is exactly the statement that $T'$ is still random.

1. Structure of Algorithm D at the root

Assume Algorithm D uses the standard deterministic rule: replace a node with two children by its inorder predecessor (the maximum key in its left subtree), then delete that predecessor from its original position.

Let

$$ b = \max L. $$

Then Algorithm D performs:

Replace the root label $a$ by $b$.
Delete the node containing $b$ from the left subtree.

Thus:

The new root is $b$,
The right subtree is unchanged,
The left subtree is the tree obtained from the original left subtree after deleting its maximum element.

So we must understand the joint distribution of:

the pair $(a,b)$,
the induced subtree structure after deleting $b$ from the left subtree.

2. Distribution of $(a,b)$

Fix distinct keys $a,b \in S$. We count permutations $\pi$ such that:

$\pi_1 = a$,
$\max{x \in L} = b$.

This means:

$b < a$,
all elements of $L$ are $< a$,
and $b$ is the largest element among those appearing after $\pi_1$ that are $< a$.

Equivalently, among the set $L = {1,\dots,a-1}$, the maximum element $a-1$ is not necessarily $b$; rather $b$ is determined by which elements appear after position $1$ in the permutation.

Now observe the key symmetry:

For fixed $a$, and fixed $b < a$, the number of permutations with $\pi_1 = a$ and with $\max L = b$ depends only on:

the size of the set ${1,\dots,a-1}$,
not on the specific value of $b$.

Indeed, relabeling the elements of ${1,\dots,a-1}$ by any order-preserving bijection that swaps two values preserves the counting condition. Hence all $b < a$ occur with equal probability.

Therefore:

$$ \mathbb{P}(\text{new root } = b \mid \pi_1 = a) = \frac{1}{n}, \quad b \in S \setminus {a}. $$

So the new root is uniformly distributed over the remaining keys.

3. Distribution of subtrees after deletion

Condition on the event $\pi_1 = a$ and $\max L = b$.

The permutation $\pi$ induces:

a random permutation of $L$ after deleting $b$,
a random permutation of $R$ unchanged,
independence between left and right parts (standard BST decomposition property).

Now apply Algorithm D inside the left subtree. The left subtree is a random BST on $L$, so deleting its maximum element $b$ produces, by the same permutation argument restricted to $L$, a random BST on $L \setminus {b}$.

Thus:

left subtree of $T'$ is a random BST on $L \setminus {b}$,
right subtree is a random BST on $R$,
and they remain independent.

4. Identification with the BST distribution

We now compare with $T(\pi')$, where $\pi'$ is a uniform permutation of $S \setminus {a}$.

In $T(\pi')$:

the root is uniformly distributed over $S \setminus {a}$,
conditioned on the root $b$, the left subtree is a random BST on ${x < b}$,
the right subtree is a random BST on ${x > b}$,
and these are independent.

We now check that $T'$ satisfies the same recursive distribution:

(i) Root distribution

We proved:

$$ \mathbb{P}(\text{root of } T' = b) = \frac{1}{n}, $$

so the root is uniform over $S \setminus {a}$.

(ii) Conditional subtree distribution

Given root $b$:

left subtree is a random BST on ${x < b}$,
right subtree is a random BST on ${x > b}$,
independence holds.

This matches exactly the defining recursive property of a random BST.

5. Conclusion

The tree $T'$ obtained by deleting the root of a random BST via Algorithm D satisfies:

uniformly random root among remaining keys,
independent random BST structure on left and right subtrees,
same recursive distribution as Algorithm T on $n$ keys.

Therefore $T'$ has the same distribution as a random binary search tree on $S \setminus {a}$.

Hence, deleting the root by Algorithm D preserves randomness.

$$ \boxed{\text{The resulting tree is still random.}} $$