Binary Search Trees
Implementation and Operations
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SoftUni Team
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Technical Trainers
Software University
http://softuni.bg
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Table of Contents
1. Binary Search Trees
2. BST Operations

Search()

DeleteMin()

Range()
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Binary Search Trees
Two Children at Most
Binary Search Trees
 Binary search trees are ordered
 For each node x
what about ==

Elements in left subtree of x are < x

Elements in right subtree of x are > x
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9
25
nodes are < 17
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11
20
31
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BST - Search
 Search for x in BST
 if node is not null
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
if x < node.value  go left

else if x > node.value  go right

else if x == node.value  return node
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6
19
12
25
Search 12  17 9 12
Search 27  17 19 25 null
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BST - Insert
 Insert x in BST
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 if node is null  insert x
 else if x < node.value  go left
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19
 else if x > node.value  go right
 else  node exists
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12
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Insert 12  17 9 12 return
Insert 27  17 19 25 null(insert)
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Problem: BST
 You are given a skeleton
 Implement BinarySearchTree<T>

bool Contains(T value)

void Insert(T value)

void EachInOrder(Action<T>)
Same as before
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1
3
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Solution: BST Node
Inside BST
private Node root;
private class Node
{
public Node(T value)
{
this.Value = value;
}
public T Value { get; set; }
public Node Left { get; set; }
public Node Right { get; set; }
}
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Solution: BST Contains
public bool Contains(T value)
{
Node current = this.root;
while (current != null)
{
if (value.CompareTo(current.Value) < 0)
current = current.Left;
else if (value.CompareTo(current.Value) > 0)
current = current.Right;
else
break;
}
return current != null;
}
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Solution: BST Insert
public void Insert(T value)
{
if (this.root == null)
{
this.root = new Node(value);
}
return;
Node parent = null;
Node current = this.root;
while(current != null) { //TODO: search for node }
//TODO: insert node
}
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Problem: BST Search
 Implement:

BST<T> Search(T value)
 Make sure the method works for:
 empty tree
 tree with one element
 tree with two elements - root + left/right
 tree with multiple elements
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Solution: BST Search
public BinarySearchTree<T> Search(T item)
{
Node current = this.root;
while (current != null)
{
if (item.CompareTo(current.Value) < 0)
current = current.Left;
else if (item.CompareTo(current.Value) > 0)
current = current.Right;
else
break;
}
return new BinarySearchTree<T>(current);
}
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Solution: BST Search (2)
private BinarySearchTree(Node root)
{
this.Copy(root);
}
private void Copy(Node node)
{
if (node == null) return;
this.Insert(node.Value);
this.Copy(node.Left);
this.Copy(node.Right);
Pre-Order
Traversal
}
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Lab Exercise
BST - Insert, Contains, Search
BST - Search Operation Speed - Quiz
TIME’S UP!
TIME:
 What is the speed of the Search(T) operation on BST?
 O(n)
 O(log(n))
 O(1)
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BST - Search Operation Speed - Answer
 What is the speed of the Search() operation on BST?
 O(n)
 O(log(n))
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 O(1)
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25
34
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BST - Remove
 Remove x from BST
 if node is null  exit
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 if node is leaf  remove
 if node is non-leaf  find replacement

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3 cases (continues…)
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18
25
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BST – Remove (1)
 1. Deleted has no right child
 promote its left child
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 Example: Delete 9
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6
19
18
25
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BST – Remove (2)
 2. Deleted right's child has no left child
 promote right child
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 Example: Delete 19
9
6
19
25
18
null
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BST – Remove (3)
 3. Deleted right's child has left child
 find min in deleted right's left
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 promote min
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19
 Example: Delete 17
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18
25
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Problem: BST Delete Min
 Implement:

void DeleteMin()
 Make sure the method works for:
 empty tree
 tree with one element
 tree with two elements - root + left/right
 tree with multiple elements
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Solution: BST Delete Min
public void DeleteMin()
{
if (this.root == null) { return; }
Node<T> parent = null;
Node<T> current = this.root;
while (current.Left != null)
{
parent = current;
current = parent.Left;
}
if (parent == null) { this.root = current.Right; }
else { parent.Left = current.Right; }
}
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Lab Exercise
BST - DeleteMin
Binary Search Trees – Operation Speed
 Insert – height of tree
 Search – height of tree
O(n)
 Delete – height of tree
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9
6
19
18
25
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Binary Search Trees – Best Case
 Example: Insert 17, 10, 25, 5, 15, 19, 34
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10
5
25
15
19
34
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Binary Search Trees – Average Case
 You can insert values in ever random order
 Example: Insert 17, 19, 9, 6, 25, 28, 18
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9
6
19
18
25
28
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Binary Search Trees – Worst Case
 You can insert values in ever increasing/decreasing order
 Example: Insert 17, 19, 25, 34
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Linked List
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25
34
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Balanced Binary Search Trees
 Binary search trees can be balanced
 Balanced trees have for each node

Nearly equal number of nodes in its subtrees
 Balanced trees have height of ~ log(n)
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BST - Range
 All elements between 2 values
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 Return enumerator with the elements
 Find the
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37
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Range 4, 37 
4 5 8 9 10 37
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4
9
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Lab Exercise
BST - Range
BST - Range Operation Speed - Quiz
TIME’S UP!
TIME:
 What is the speed of the Range(T, T) operation on BST?
 O(n)
 O(log(n))
 O(1)
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BST - Range Operation Speed - Answer
 What is the speed of the Range(T, T) operation on BST?
 O(n)
 O(log(n))
 O(1)
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Summary
 Binary search trees are ordered binary trees
 Balanced trees have roughly the same height
of their left and right children
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Binary Search Trees
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
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
"Data Structures and Algorithms" course by Telerik Academy under CC-BY-NC-SA license
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