1. Binary Search Tree
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2. Outline Concept of Binary Search Tree (BST) BST operations
Find
Insert
Remove
Running time analysis of BST operations
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3. Binary Search Tree: Properties
Elements have keys (no duplicates allowed).
For every node X in the tree, the values of all the keys in the left subtree are smaller than the key in X and the values of all the keys in the right subtree are larger than the key in X.
The keys must be comparable. X
<X
>X
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4. Binary Search Tree: Examples7 9 2 1 5 6 3
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5. Binary Search Tree: Examples3 3 1 1 2 2 1 3 3 2 1 2 2 1 3
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6. Basic Operations FindMin, FindMax, Find Insert Remove
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7. FindMin Find node with the smallest value Algorithm: Code:
Keep going left until you reach a dead end!
Code:
BinaryNode<Type> findMin(BinaryNode<Type> t) {
if (t != null)
while (t.left != null)
t = t.left;
return t; }
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8. FindMax Find node with the largest value Algorithm: Code:
Keep going right until you reach a dead end!
Code:
BinaryNode<Type> findMax(BinaryNode<Type> t) {
if (t != null)
while (t.right != null)
t = t.right;
return t; }
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9. Find
You are given an element to find in a BST. If it exists, return the node. If not, return null.
Algorithm?
Code? 7 9 2 1 5 6 3
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10. Find: Implementation
BinaryNode<Type> find(Type x, BinaryNode<T> t) { while(t!=null) if(x.compareTo(t.element)<0) t = t.left; else if(x.compareTo(t.element)>0) t = t.right; else return t; // Match } return null; // Not found
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11. Insertion: Principle
When inserting a new element into a binary search tree, it will always become a leaf node. 10 2 3 15 1 5 6 12 14
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12. Insertion: Algorithm To insert X into a binary search tree:
Start from the root
If the value of X < the value of the root: X should be inserted in the left sub-tree.
If the value of X > the value of the root: X should be inserted in the right sub-tree.
Remember that a sub-tree is also a tree.
We can implement this recursively!
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13. Insertion: Implementation
BinaryNode<Type> insert(Type x, BinaryNode<Type> t) { if (t == null) t = new BinaryNode<Type>(x); else if(x.compareTo(t.element)<0) t.left = insert (x, t.left); else if(x.compareTo(t.element)>0) t.right = insert (x, t.right); else throw new DuplicateItemException(x); return t; }
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14. Removing An Element 8 4 5 12 1 6 3 4 6 5
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15. Removing An Element: Algorithm
If the node is a leaf, simply delete it.
If the node has one child, adjust parent’s child reference to bypass the node.
If the node has two children:
Replace the node’s element with the smallest element in the right subtree and then remove that node, or
Replace the node’s element with the largest element in the left subtree and then remove that node
Introduces new sub-problems:
removeMin:
Alternatively, removeMax
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16. Removing Leaf 8 12 4 6 1 3 5
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17. Removing Node With 1 Child8 12 4 6 1 3 5
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18. Removing Node With 1 Child8 12 4 6 1 3 5
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19. removeMin
BinaryNode<Type> removeMin(BinaryNode<Type> t) { if (t == null) throw new ItemNotFoundException(); else if (t.left != null) t.left = removeMin(t.left); return t; } else return t.right;
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20. Removing Node With 2 Children7 9 2 1 5 3 4
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21. Removing Node With 2 Children7 2 9 3 3 1 5 3 4
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22. Removing Node With 2 Children7 2 9 3 2 1 5 3 4
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23. Removing Root 7 2 3 12 1 5 4 10 14 9 11 9
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24. Remove
BinaryNode<Type> remove(Type x, BinaryNode<Type> t) { if (t == null) throw new ItemNotFoundException(); if (x.compareTo(t.element)<0) t.left = remove(x, t.left); else if(x.compareTo(t.element)>0) t.right = remove(x, t.right); else if (t.left!=null && t.right != null) t.element = findMin(t.right).element; t.right = removeMin(t.right); } else if(t.left!=null) t=t.left; else t=t.right; return t;
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25. Find k-th element X X X SL SR SL SR SL SR k < SL + 1 k == SL + 1
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26. Find k-th element
BinaryNode<Type> findKth(int k, BinaryNode<Type> t) { if (t == null) throw exception; int leftSize = (t.left != null) ? t.left.size : 0; if (k <= leftSize ) return findKth (k, t.left); else if (k == leftSize + 1) return t; else return findKth ( k - leftSize - 1, t.right); }
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27. Analysis Running time for: Average case: O(log n) Worst case: O(n)
Insert?
Find min?
Remove?
Find?
Average case: O(log n)
Worst case: O(n)
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28. Summary Binary Search Tree maintains the order of the tree.
Each node should be comparable
All operations take O(log n) - average case, when the tree is equally balanced.
All operations will take O(n) - worst case, when the height of the tree equals the number of nodes.
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