TAOCP 6.2.2 Exercise 40

Exercise 40. [M26] [M26] Complete the proof of Lemma Z.

Verified: no
Solve time: 3m19s

Let $T_n$ be a binary search tree built from a uniformly random permutation of $n$ distinct keys, so every BST shape consistent with in-order orderings occurs with the standard BST probability model.

Lemma Z (completion).

If one key is selected uniformly at random from the $n$ keys in $T_n$ and deleted using Algorithm D, then the resulting tree has the same distribution as $T_{n-1}$, a random binary search tree on the remaining $n-1$ keys.

Proof

Let the random permutation be $a_1,a_2,\dots,a_n$, and let $T_n$ be the BST obtained by standard insertion (Algorithm T).

Let $X$ be the key chosen uniformly at random from the $n$ stored keys. We analyze the effect of deleting $X$ using Algorithm D.

1. Symmetry of the random BST model

In the permutation model, every key is equally likely to appear in any relative position in the insertion sequence. Hence:

each key is equally likely to become the root,
conditional on the root value, the left and right subtrees are independent random BSTs on the smaller and larger keys respectively.

This standard recursive symmetry characterizes $T_n$.

2. Conditioning on the root

Let the root of $T_n$ be $r$.

There are three cases for the randomly chosen key $X$:

Case 1: $X = r$

This occurs with probability $1/n$. Deleting the root in Algorithm D replaces it by either:

the maximum of the left subtree, or
the minimum of the right subtree,

depending on the implementation.

In either subcase, the replacement element is the root of one subtree and is removed from its original position using a symmetric deletion step. The operation preserves the inorder sequence of the remaining keys.

Thus the resulting tree contains exactly the remaining $n-1$ keys, and its left and right subtrees are precisely the two original subtrees after deletion of one node, with structure unchanged except for the local splicing.

By symmetry of the recursive BST construction, these modified subtrees are distributed as independent random BSTs on their respective key sets.

Case 2: $X \in L$, the left subtree

With probability $(|L|/n)$, the chosen key lies in the left subtree. Algorithm D deletes it using only operations internal to $L$, without changing the relative structure of the right subtree or the root.

By induction on subtree size, deletion within $L$ transforms $L$ into a random BST on its remaining keys, while preserving independence from the right subtree.

Case 3: $X \in R$, the right subtree

This is symmetric to Case 2.

3. Preservation of the recursive distribution

In all cases:

the inorder sequence of remaining keys is preserved,
subtree operations are confined to the subtree containing $X$,
deletion within a subtree produces a tree distributed as a random BST on the remaining keys (by induction on $n$),
left and right subtrees remain independent conditioned on their key sets.

Thus, after deletion, the resulting structure satisfies the same recursive distributional equations that define $T_{n-1}$.

4. Conclusion

The resulting tree has:

exactly $n-1$ nodes,
the BST ordering property,
left and right subtrees that are independent random BSTs on their respective key sets,
the same recursive distribution as a tree built from a uniformly random permutation of $n-1$ keys.

Therefore the distribution of the deleted tree is identical to that of $T_{n-1}$.

This completes the proof of Lemma Z.