TAOCP 6.2.2 Exercise 48

Section 6.2.2: Binary Tree Searching

Exercise 48. [**] [M47| Carry out a rigorous analysis of the steady state of a binary search tree subjected to random insertions and deletions.

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Let $T_n$ denote the set of binary search trees on $n$ distinct keys, and consider the Markov process in which at each step an insertion of a random key and a deletion of a uniformly chosen node are performed, with both operations implemented by Algorithm T and Algorithm D.

By the construction in Theorem H, if a tree on $n$ nodes is produced by inserting keys in random order, then deleting a uniformly chosen element by Algorithm D yields a tree distributed exactly as if it had been produced by a random insertion sequence of length $n$ followed by relabeling. Hence the uniform distribution on permutations of $n$ keys induces a stationary distribution on $T_n$ under deletion, since deletion corresponds to a uniform random removal at the permutation level without bias in the induced tree structure.

Insertion of a random key corresponds to inserting a uniformly random element into a uniformly random permutation and applying Algorithm T, which preserves the uniform permutation model with one additional element. The joint effect of insertion followed by deletion therefore leaves the induced permutation on the current $n$ keys uniformly distributed over all $n!$ possibilities. Consequently the induced distribution on tree shapes is invariant under the combined operations, and the process is ergodic on the finite state space $T_n$.

The stationary regime is therefore the same as that of a random binary search tree obtained from a uniformly random permutation, so all asymptotic properties derived for random insertion trees apply unchanged in steady state, including average successful search cost $2H_n+O(1) \sim 2\ln n$. This completes the proof. ∎