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A computer scientist whose work with Robert Tarjan includes splay trees and amortized analysis. His Carnegie Mellon University homepage identifies his full name and academic role.^[ref]

First perform splaying on the key to be deleted, making its node the root of a tree. Delete that root, leaving left and right subtrees $L,R$. If either is empty, return the other. Otherwise find the maximum key $m$ of $L$ by following right-child references, and perform splaying on $m$ within $L$. This makes $m$ the root of $L$ with an empty right subtree. Attach $R$ there. The result is the original splay tree with exactly the selected key removed. An unsuccessful lookup leaves the represented set unchanged.

For algorithm correctness, all keys of $L$ are smaller than the deleted key, and all keys of $R$ are larger. Splaying preserves the binary search tree property by the permitted assumption. Since $m$ is the maximum of $L$, it has no larger key within $L$; since every key of $R$ exceeds $m$, attaching $R$ as its right subtree preserves every ordering condition. All remaining keys occur once. This is the join-based deletion of a splay tree.^[ref]

Use the usual bottom-up splaying operations. We prove the stronger result that at most three successful calls suffice. No insertions or deletions are needed.

If the chosen key $x$ is the minimum stored key, first perform splaying on $x$ and then on its immediate successor. The argument for consecutive keys in a binary search tree makes $x$ the left child of the successor with an empty right subtree; minimality makes its left subtree empty too. Hence $x$ is a leaf of a rooted tree. The maximum-key case is the mirror image.

Suppose $x$ is an interior key. Since at least four keys are stored, there are either two keys larger than $x$ or two smaller than $x$. In the former case choose four consecutive stored keyswhere $a$ is the predecessor of $x$, and $b,c$ are the next two successors. Perform $\mathrm{Splay}(c),\mathrm{Splay}(a),\mathrm{Splay}(b)$.

Here is a complete check of the local shapes. After the first call, $c$ is the root. After the second, $a$ is the root; every key below $a$ is in its left subtree $A$, and every key above $c$ stays in the right subtree $D$ of $c$. These outside subtrees cannot interfere with the next search. Between $a$ and $c$ there are only $x,b$. The last step of splaying on $a$ is either a zig step, leaving $c$ as its right child; a zig-zag step, with the same result; or a zig-zig step, leaving some $p\in \{x,b\}$ as its right child with $c$ immediately to the right of $p$. Consequently the possible right subtrees of $a$ are exactly among these four forms, using $k(L,R)$ for a node with its left and right subtrees:Here an unexpanded $x$ or $b$ denotes a node with no children. Each row follows by the indicated one or two tree rotations in each bottom-up step. All rows end with $x$ as the right child of $a$, with both its subtrees empty. If instead two keys below $x$ are available, reflect the construction, exchanging left and right and reversing the order of keys. This proves splaying a chosen node into a leaf for every node of every splay tree of size at least four.

Original article in Journal of the ACM 32(3), pp. 652–686. Author-hosted publication record and abstract: www.cs.cmu.edu/~sleator/papers/Self-Adjusting.htm

The article introduces splay trees and proves a logarithmic amortized time complexity for their standard operations.

Every node of a splay tree with at least four distinct keys can become a leaf of a rooted tree using only successful splaying operations. At an extreme key, access that key and then its sole adjacent stored key; the consecutive keys in a binary search tree property makes the extreme key a leaf. For an interior key $x$ with two larger stored keys, take consecutive $a<x<b<c$ and access $c,a,b$; the mirror construction handles two smaller keys.

For a quick shape proof, use $k(L,R)$ for a node and its two subtrees. After accessing $c,a$, the right subtree of $a$ can only be $c(b(x,0),D)$, $c(x(0,b),D)$, $b(x,c(0,D))$, or $x(0,c(b,D))$, with outside subtrees $A$ below $a$ and $D$ above $c$. This follows by inspecting the final zig step, zig-zig step or zig-zag step that moves $a$ above the former root $c$. In each case, accessing $b$ yields $b(a(A,x),c(0,D))$, where $x$ is a leaf and $0$ denotes an empty subtree. These rotations prove the bound for arbitrary outside subtrees.

A splay tree is a binary search tree whose access operations use splaying to move a selected key to the root through tree rotations. It stores no explicit balance or height information. A single access can take linear time, while the standard operations have logarithmic amortized time complexity. The distinction matters: amortized analysis bounds the total cost of an operation sequence, not the worst cost of every individual operation.^[ref]