Binary search tree: find: Difference between revisions
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'''Variant:''' <math>i</math> is increased by 1. | '''Variant:''' <math>i</math> is increased by 1. | ||
'''Break condition:''' Either it is <math>p =</math>void or, otherwise, it is<math>p</math>.key <math>= K</math>. | '''Break condition:''' Either it is <math>p =</math>void or, otherwise, it is <math>p</math>.key <math>= K</math>. | ||
== Induction basis == | == Induction basis == |
Revision as of 07:33, 18 May 2015
General Information
Algorithmic Problem: Sorted Sequence:find
Type of algorithm: loop
Auxiliary data: A pointer [math]\displaystyle{ p }[/math] of type "pointer to binary search tree node of type [math]\displaystyle{ \mathcal{K} }[/math]."
Abstract view
Invariant: After [math]\displaystyle{ i\geq 0 }[/math] Iterations.
- The pointer [math]\displaystyle{ p }[/math] points to a tree node [math]\displaystyle{ v }[/math] on height level [math]\displaystyle{ i }[/math] (or is void).
- The key [math]\displaystyle{ K }[/math] is in the range of [math]\displaystyle{ v }[/math].
Variant: [math]\displaystyle{ i }[/math] is increased by 1.
Break condition: Either it is [math]\displaystyle{ p = }[/math]void or, otherwise, it is [math]\displaystyle{ p }[/math].key [math]\displaystyle{ = K }[/math].
Induction basis
Abstract view: Set [math]\displaystyle{ p:= }[/math] root.
Implementation: Obvious
Proof: Nothing to show
Induction step
Abstract view: If p points to a node but not with key K, p descends in the appropriate direction, left or right.
Implementation:
- If [math]\displaystyle{ p = }[/math]void, terminate the algorithm and return 'false'.
- Otherwise, if [math]\displaystyle{ p }[/math].key [math]\displaystyle{ = K }[/math], terminate the algorithm and return 'true'.
- Otherwise:
- If [math]\displaystyle{ K \lt p }[/math].key, set [math]\displaystyle{ p := }[/math]left.
- If [math]\displaystyle{ K \gt p }[/math].key, set [math]\displaystyle{ 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.
Pseudocode
TREE-SEARCH (x, k)
- if x= NIL or k = key[x]
- then return x
- if k < key[x]
- then return TREE-SEARCH(left[x], k)
- else return TREE-SEARCH(right[x], k)