Binary Search Tree
Sorting and Searching in a Dynamic Set
S.V. N. (vishy) Vishwanathan
University of California, Santa Cruz
[email protected]
February 29, 2016
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Binary Search Trees
Outline
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Binary Search Trees
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Binary Search Trees
What is a Binary Search Tree?
Binary tree (each node has 0, 1, or 2 children)
Each node contains
key
satellite data
left
right
parent
The keys satisfy the binary search tree property
Let x be a node in a binary search tree. If y is a node in the
left subtree of x, then y .key ≤ x.key . If y is a node in the
right subtree of x, then y .key ≥ x.key .
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Binary Search Trees
Examples
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Binary Search Trees
Inorder Tree Walk
inorder-tree-walk(x)
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if x 6= nil
inorder-tree-walk(x.left)
print x.key
inorder-tree-walk(x.right)
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Binary Search Trees
Search
tree-search(x, k)
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if x == nil or k == x.key
return x
if k < x.key
return tree-search(x.left, k)
else return tree-search(x.right, k)
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Binary Search Trees
Iterative Search
iterative-tree-search(x, k)
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while x 6= nil and k 6= x.key
if k < x.key
x = x.left
else x = x.right
return x
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Binary Search Trees
Search
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Binary Search Trees
Minimum
tree-minimum(x)
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while x.left 6= nil
x = x.left
return x
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Binary Search Trees
Maximum
tree-maximum(x)
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while x.right 6= nil
x = x.right
return x
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Binary Search Trees
Minimum and Maximum
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Binary Search Trees
Successor
Given a node x, its successor is the node with the smallest key greater
than x.key
It is the next node which occurs after x in the sorted order
determined by an inorder tree walk
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Binary Search Trees
Successor
Look at node 15 and 13
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Binary Search Trees
Successor
tree-successor(x)
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if x.right 6= nil
return tree-minimum(x.right)
y = x.parent
while y 6= nil and x == y .right
x =y
y = y .parent
return y
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Binary Search Trees
Time Complexity
Theorem
We can implement the dynamic-set operations search, minimum,
maximum, successor, and predecessor so that each one runs in
O(h) time on a binary search tree of height h
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Binary Search Trees
Insertion
tree-insert(T , z)
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y = nil
x = T .root
while x 6= nil
y =x
if z.key < x.key
x = x.left
else x = x.right
z.parent = y
if y == nil
T .root = z
// tree T was empty
elseif z.key < y .key
y .left = z
else y .right = z
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Binary Search Trees
Insertion
Look at node 13
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Binary Search Trees
Deletion: Three Cases
z has no children. Delete z and modify its parent
z has one child. Elevate child to take z’s position
z has two children.
Find z’s successor y . We know that
y has no left child. Why?
y must lie in the right subtree of z
Replace z with y
However have to be careful if y is the right child of z
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Binary Search Trees
Deletion
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Binary Search Trees
Deletion: Split to Four Cases
z has no left child. Replace z by its right child (which may be a nil
z has a left child. Replace z by its left child.
z has both left and right children.
Find z’s successor y
y must lie in the right subtree of z and has no left child
If y is z’s right child, then replace z by y
Otherwise, replace y by its right child, and replace z by y
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Binary Search Trees
Pseudo-code
tree-delete(T , z)
1 if z.left == nil
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transplant(T , z, z.right)
3 elseif z.right == nil
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transplant(T , z, z.left)
5 else y = tree-minimum(z.right)
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if y .parent 6= z
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transplant(T , y , y .right)
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y .right = z.right
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y .right.parent = y
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transplant(T , z, y )
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y .left = z.left
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y .left.parent = y
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Binary Search Trees
Pseudo-code
transplant(T , u, v )
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if u.parent == nil
T .root = v
elseif u == u.parent.left
u.parent.leftgetsv
else u.parent.right = v
if v 6= nil
v .parent = u.parent
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Binary Search Trees
Questions?
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