Trees 3: The Binary Search Tree
• Section 4.3
1
Binary Search Tree
• A binary tree B is called a binary search tree iff:
– There is an order relation < defined for the vertices of B
– For any vertex v, and any descendant u in the subtree v.left,
u<v
– For any vertex v, and any descendent w in the subtree
v.right, v < w
root
1
4
2
6
3
5
7
2
Binary Search Tree
Which one is NOT a BST?
3
Binary Search Tree
• Consequences
– The smallest element in a binary search tree
(BST) is the “left-most” node
– The largest element in a BST is the “right-most”
node
– Inorder traversal of a BST encounters nodes in
increasing order
root
1
4
2
6
3
5
7
4
Binary Search using BST
• Assumes nodes are organized in a binary search tree
– Begin at root node
– Descend using comparison to make left/right
decision
• if (search_value < node_value) go to the left child
• else if (search_value > node_value) go to the right child
• else return true (success)
– Until descending move is impossible
– Return false (failure)
5
Binary Search using BST
• Runtime <= descending path length <= depth of tree
• If tree has “enough” branching, runtime is O(log n)
– Worst case is O(n)
6
BST Class Template
7
BST Class Template (contd.)
Pointer passed by reference
(why?)
Internal functions
used in recursive
calls
8
BST: Public members calling private recursive
functions
9
BST: Searching for an element
10
BST: Find the smallest element
Tail recursion
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BST: Find the biggest element
Non-recursive
12
BST: Insertion (5)
Before insertion
After insertion
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BST: Insertion (contd.)
Strategy:
•Traverse the tree
as in searching for t
with contains()
•Insert if you cannot
find t
14
BST: Deletion of Leaf
Deleting a node with no child
Before deleting (3)
After deleting (3)
15
Deletion Strategy: Delete the node
BST: Delete a Node with One Child
Deleting a node with one child
Before deleting (4)
After deleting (4)
Deletion Strategy: Bypass the node being deleted 16
BST: Delete a Node with Two Children
Deleting a node with two children
Before deleting (2)
After deleting (2)
Replace the node with smallest node in the right subtree
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BST: Deletion Code
18
BST Deletion
2
3
Element: 5
5
Left: 160
208 Right: 0
3
Address 208
Element: 3
4
Left: 0 Right: 160
Address 160
Element: 4
Left: 0 Right: 0
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BST: Lazy Deletion
• Another deletion strategy
– Don’t delete!
– Just mark the node as deleted
– Wastes space
– But useful if deletions are rare or space is not a
concern
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BST: Insertion Bias
•
•
•
•
Start with an empty tree
Insert elements in sorted order
What tree do you get?
How do you fix it?
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BST: Deletion Bias
After large number of alternating insertions and deletions
Why this bias? How do you fix it?
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BST: Search using function objects
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Average Search/Insert Time - 1
• Average time is the average depth of a vertex
– Let us compute the sum of the depths of all vertices and
divide by the number of vertices
– The sum of the depths is called the internal path length
• Give the internal path lengths for the following trees
2
0
2
1
6
2
3
9
5
1
2
3
8
7
4
6
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Average Search/Insert Time - 2
Let D(N) be the internal path length of a tree with N vertices
If the root has a left subtree with i nodes, then
•D(N) = D(i) + D(N-i-1) + N-1
because the depth of each vertex in the subtrees increases by 1
6
7
2
1
2
9
5
2
7
1
5
9
8
11
8
3
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Average Search/Insert Time - 3
The average value of D(N) is given by the recurrence
•D(1) = 0
•D(N) = 1/N[ i=0N-1 D(i) + D(N-i-1)] + N - 1
•
= 2/N i=0N-1 D(i) + N - 1
Root
Subtree
with N-1
nodes
Root
Subtree
with 1
node
Subtree
with N-2
nodes
Root
Subtree
with 2
nodes
Subtree
with N-3
nodes
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Average Search/Insert Time - 4
•D(N) = 2/N i=0N-1 D(i) + N - 1
•N D(N) = 2 i=0N-1 D(i) + N(N - 1)
•(N-1)D(N-1) = 2 i=0N-2 D(i) + (N-1)(N - 2)
(2) - (1) gives
•ND(N) - (N-1)D(N-1) = 2D(N-1) + 2(N-1)
•ND(N) = (N+1)D(N-1) + 2(N-1)
•D(N)/(N+1) = D(N-1)/N + 2(N-1)/[N(N+1)]
•
< D(N-1)/N + 2/N
(1)
(2)
•D(N)/(N+1) < D(N-1)/N + 2/N
•D(N-1)/(N) < D(N-2)/(N-1) + 2/(N-1)
•D(N-2)/(N-1) < D(N-3)/(N-2) + 2/(N-2)
•...
•D(2)/(3) < D(1)/2 + 2/2
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Average Search/Insert Time - 5
•D(N)/(N+1) < D(N-1)/N + 2/N
•D(N-1)/(N) < D(N-2)/(N-1) + 2/(N-1)
•D(N-2)/(N-1) < D(N-3)/(N-2) + 2/(N-2)
•...
•D(2)/(3) < D(1)/2 + 2/2
•D(N)/(N+1) < D(N-1)/N + 2/N
•
< D(N-2)/(N-1) + 2/(N-1) + 2/N
•
< D(N-3)/(N-2) + 2/(N-2) + 2/(N-1) + 2/N
•...
•
< D(1)/(2) + 2/2 + ... + 2/(N-2) + 2/(N-1) + 2/N
•
= 2 i=2N 1/i
If we show that i=2N 1/i is O(log N), then we can prove that average
D(N) = O(N Log N) and so the average depth is O(log N)
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Deriving Time Complexity Using Integration
•Area under the rectangles is smaller than that under 1/x
• i=24 1/i < ∫14 1/x dx
N
• i=2N 1/i < ∫1 1/x dx = ln (N) - ln (1) = O(log N)
f(x) = 1/x
1/
2
1
1/
3
2
1/
4
3
4
•Integration can be used to derive good bounds for sums of the form
i=aN f(i) when f(i) is monotonically increasing or decreasing if you know
how to integrate f(x)
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