Binary search tree: remove node

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General Information

Algorithmic problem: See the remark clause of Binary Search Tree; pointer [math]\displaystyle{ p }[/math] as defined there is the input.

Prerequisites: [math]\displaystyle{ p }[/math].left [math]\displaystyle{ \neq }[/math] void.

Type of algorithm: loop

Auxiliary data: A pointer [math]\displaystyle{ p' }[/math] of type "pointer to a binary search tree node of type [math]\displaystyle{ \mathcal{K} }[/math]."

Abstract View

Invariant:

  1. The immediate predecessor of K is in the range of [math]\displaystyle{ p' }[/math].
  2. It is [math]\displaystyle{ p' }[/math].right [math]\displaystyle{ \neq }[/math] void.

Variant: The pointer [math]\displaystyle{ p' }[/math] descends one level deeper, namely to [math]\displaystyle{ p }[/math]'.right.

Break condition: It is [math]\displaystyle{ p' }[/math].right.right = void.

Remark: For example, the height of the subtree rooted at the node to which [math]\displaystyle{ p }[/math] points may be chosen as the induction parameter. For conciseness, the induction parameter is omitted in the following.

Induction Basis

Abstract view: If [math]\displaystyle{ p.left }[/math] is the immediate predecessor of K, overwrite K by its immediate predecessor and terminate; otherwise, initialize [math]\displaystyle{ p' }[/math].

Implementation:

  1. If [math]\displaystyle{ p }[/math].left.right = void:
    1. Set [math]\displaystyle{ p }[/math].key := p.left.key.
    2. Set [math]\displaystyle{ p }[/math].left := p.left.left.
    3. Terminate the algorithm.
  2. Otherwise, set [math]\displaystyle{ p' }[/math] := p.left.

Proof: Obvious.

Induction Step

Abstract view: If [math]\displaystyle{ p' }[/math].right.key is the immediate predecessor of [math]\displaystyle{ K }[/math], overwrite [math]\displaystyle{ K }[/math] by its immediate predecessor and terminate; otherwise, let [math]\displaystyle{ p' }[/math] descend one level deeper.

Implementation:

  1. If [math]\displaystyle{ p' }[/math].right.right = void:
    1. Set [math]\displaystyle{ p }[/math].key := [math]\displaystyle{ p' }[/math].right.key.
    2. Set [math]\displaystyle{ p' }[/math].right := [math]\displaystyle{ p' }[/math].right.left.
    3. Terminate the algorithm.
  2. Set [math]\displaystyle{ p':=p' }[/math].right.

Correctness: Obvious.

Complexity

Statement: The complexity is in [math]\displaystyle{ \mathcal{O}(T\cdot h)\subseteq\mathcal{O}(T\cdot n) }[/math] in the worst case, where [math]\displaystyle{ n }[/math] is the length of the sequence, [math]\displaystyle{ h }[/math] the height of the tree, and [math]\displaystyle{ T }[/math] the complexity of the comparison.

Proof: Obvious.