Binary search tree: insert

Binary Search Tree
Insert

Sorted sequence

Openlearnware
See Chapter 4

General Information

Algorithmic problem: Sorted sequence: insert

Type of algorithm: loop

Auxiliary data: A pointer p 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:

1. The pointer [math]\displaystyle{ p }[/math] points to a tree node [math]\displaystyle{ v }[/math] on height level [math]\displaystyle{ i }[/math].
2. The Key K is in the range of v.

Variant: [math]\displaystyle{ i }[/math] increased by 1.

Break condition: One of the following two conditions is fulfilled:

1. It is [math]\displaystyle{ p }[/math].key [math]\displaystyle{ \geq K }[/math] and [math]\displaystyle{ p }[/math].left [math]\displaystyle{ = void }[/math].
2. It is [math]\displaystyle{ p }[/math].key [math]\displaystyle{ \leq K }[/math] and [math]\displaystyle{ p }[/math].right [math]\displaystyle{ = }[/math]void.

Induction Basis

Abstract view: If the tree is empty, a new root with key K is created; otherwise, p is initialized so as to point to the root.

Implementation:

1. If [math]\displaystyle{ root = void }[/math]:
   1. Create a new binary search tree node and let root point to it.
   2. Set [math]\displaystyle{ root.key := K, root.left := void }[/math]. and [math]\displaystyle{ root.right := void }[/math].
2. Otherwise, set [math]\displaystyle{ p := root }[/math].

Proof: Obvious.

Induction Step

Abstract view: If the direction where to go next is void, insert K in that empty slot and terminate the algorithm. Otherwise, proceed in that direction.

Implementation:

1. If [math]\displaystyle{ p.key = K }[/math]:
   1. If [math]\displaystyle{ p.left = void }[/math]:
      1. Create a new node and let [math]\displaystyle{ p.left }[/math] point to it.
      2. Set [math]\displaystyle{ p.left.key := K }[/math], [math]\displaystyle{ p.left.left := void }[/math], and [math]\displaystyle{ p.left.right := void }[/math].
      3. Terminate the algorithm.
   2. Otherwise, if [math]\displaystyle{ p.right = void }[/math]:
      1. Create a new node and let [math]\displaystyle{ p.right }[/math] point to it.
      2. Set [math]\displaystyle{ p.left.key := K }[/math], [math]\displaystyle{ p.left.left := void }[/math], and [math]\displaystyle{ p.left.right := void }[/math].
      3. Terminate the algorithm.
   3. Otherwise, set [math]\displaystyle{ p := p.left }[/math].
2. Otherwise, if [math]\displaystyle{ p.key \gt K }[/math]:
   1. If [math]\displaystyle{ p.left = void }[/math]:
      1. Create a new node and let [math]\displaystyle{ p.left }[/math] point to it.
      2. Set [math]\displaystyle{ p.left.key := K }[/math], [math]\displaystyle{ p.left.left := void }[/math], and [math]\displaystyle{ p.left.right := void }[/math].
      3. Terminate the algorithm.
   2. Otherwise, set [math]\displaystyle{ p := p.left }[/math].
3. Otherwise (that is, [math]\displaystyle{ p.key \lt K }[/math]):
   1. If [math]\displaystyle{ p.right = void }[/math]:
      1. Create a new node and let [math]\displaystyle{ p.right }[/math] point to it.
      2. Set [math]\displaystyle{ p.right.key := K }[/math], [math]\displaystyle{ p.right.left := void }[/math], and [math]\displaystyle{ p.right.right := void }[/math].
      3. Terminate the algorithm.
   2. Otherwise, set [math]\displaystyle{ p := p.right }[/math].

Complexity

Statement: Linear in the length of the sequence in the worst case (more precisely, linear in the height of the tree).

Worst case binary search tree

Proof: Obvious.

Pseudocode

TREE-INSERT(T, z)

y = Null

x = root(T)

while x ≠ NULL

y = x

if key[z] < key[x]

then x = left[x]

then x = right[x]

p[z] = y

if y = NULL

then root[T] = z //Tree was empty

else if key[z] < key[y]

then left[y] = z

else right[y] = z