Binary search trees Search Insertion Removal February 14, 2020 - - PowerPoint PPT Presentation

Binary search trees

Search Insertion Removal

February 14, 2020 Cinda Heeren / Andy Roth / Geoffrey Tien 1

Motivation for an efficient lookup

• Stores values associated with user-specified keys

– Values may be any (homogenous) type – Keys may be any (homogenous) comparable type

• Dictionary operations

– Create – Destroy – Insert – Find – Remove

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Dictionary ADT

• Super 9 LC
• Smell like a lawnmower
• Z125 Pro
• Fun in the sun!
• CB300F
• For the mild-mannered

commuter

Insert

• Feet
• Useful for something,

presumably

Find(Z125 Pro)

• Z125 Pro
• Fun in the sun!

Data structures for Dictionary ADT

Search Insert Remove

• Linked list
• Unsorted array
• Sorted array
• Ordered linked list

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Naïve implementations, complexity

BST search

• To find a value in a BST search from the root node:

– If the target is less than the value in the node search its left subtree – If the target is greater than the value in the node search its right subtree – Otherwise return true, (or a pointer to the data, or …)

• How many comparisons?

– One for each node on the path – Worst case: height of the tree + 1

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BST search example

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17 27 search(27);

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BST search example

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17 13 16 search(16);

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BST search example

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17 13 9 11 search(12);

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BST insertion

• The BST property must hold after insertion
• Therefore the new node must be inserted in the correct

position

– This position is found by performing a search – If the search ends at the (null) left child of a node make its left child refer to the new node – If the search ends at the right child of a node make its right child refer to the new node

• The cost is about the same as the cost for the search algorithm,

O(height)

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BST insertion example

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47 32 63 19 10 23 41 37 44 54 53 59 79 96 7 12 30 43 57 91 97 Insert 43 Create new node 43 Find position Link node

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BST insertion example

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Create new BST Insert 3 Insert 15 Insert 21 Insert 23 Insert 37 3 15 21 23 37 Search 45 How many operations for Search? Complexity?

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Find min, Find max

• Find minimum:

– From the root, keep following left child links until no more left child exists

• Find maximum:

– From the root, follow right child links until no more right child exists

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43 18 68 12 7 9 33 52 56 67 27 39 50 21

BST removal

• After removal the BST property must hold
• Removal is not as straightforward as search or insertion

– With insertion the strategy is to insert a new leaf – Which avoids changing the internal structure of the tree – This is not possible with removal

• Since the removed node's position is not chosen by the algorithm
• There are a number of different cases that must be considered

– But first, let's look at an easy (lazy way)

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BST removal cases

• The node to be removed has no children

– Remove it (assigning null to its parent’s reference)

• The node to be removed has one child

– Replace the node with its subtree

• The node to be removed has two children

– …??

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Proper BST removal

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BST removal

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Target is a leaf node (no children)

47 32 63 19 10 23 41 37 44 54 53 59 79 96 7 12 30 57 91 97 Remove 30

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BST removal

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Target has one child

47 32 63 19 10 23 41 37 44 54 53 59 79 96 7 12 30 57 91 97 Remove 79 Replace with subtree

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Looking at the next node

• One of the issues with implementing a BST is the necessity to

look at both children

– And, just like a linked list, look ahead for insertion and removal – And check that a node is null before accessing its member variables

• Consider removing a node with one child in more detail

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Looking ahead

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63 54 53 59 79 96 57 91 97 Remove 59 Step 1: We need to find the node to remove and its parent To make the correct link, we need to know if the node to be removed is a left or right child

while (nd->data != target) { if (nd == NULL) return; if (target < nd->data) { parent = nd; nd = nd->left; isLeftChild = true; } else { parent = nd; nd = nd->right; isLeftChild = false; } }

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Left or right?

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63 54 53 59 79 96 57 91 97 Remove 59

while (nd->data != target) { if (nd == NULL) return; if (target < nd->data) { parent = nd; nd = nd->left; isLeftChild = true; } else { parent = nd; nd = nd->right; isLeftChild = false; } }

nd parent nd isLeftChild = true parent nd false Now we have enough information to detach 59, after attaching its child to 54.

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Removing a node with 2 children

• The most difficult case is when the node to be removed has

two children

– The strategy when the removed node had one child was to replace it with its child – But when the node has two children problems arise

• Which child should we replace the node with?

– We could solve this by just picking one …

• But what if the node we replace it with also has two children?

– This will cause a problem

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Removed node has 2 children

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47 32 63 19 10 23 41 37 44 54 53 59 79 96 7 12 30 57 91 97 Remove 32 Let’s try replacing it with one of its subtrees like the 1-child case e.g. 41 But 41 already has 2 children and must “adopt” the other child of 32 Cannot have 3 children

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Find the predecessor

• When a node has two children, instead of replacing it with one
• f its children find its predecessor

– A node’s predecessor is the right most node of its left subtree – The predecessor is the node in the tree with the largest value less than the node’s value

• The predecessor cannot have a right child and can therefore

have at most one child

– Why?

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Predecessor node

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57 59 54 53 44 41 37 30 12 7 10 23 19 32 47 63 79 96 91 97 32’s predecessor The predecessor of 32 is the rightmost node in its left subtree The predecessor cannot have a right child as then it would not be the predecessor

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Why use the predecessor?

• The predecessor has some useful properties

– Because of the BST property it must be the largest value less than its ancestor’s value

• It is to the right of all of the nodes in its ancestor’s left subtree so must be

greater than them

• It is less than the nodes in its ancestor’s right subtree

– It can have at most only one child

• These properties make it a good candidate to replace its

ancestor

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What about the successor?

• The successor to a node is the left most child of its right

subtree

– It has the smallest value greater than its ancestor’s value – And cannot have a left child

• The successor can also be used to replace a removed node

– Pick either one, but be consistent!

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Removed node has 2 children

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Replacement with successor

57 59 54 53 44 41 37 30 12 7 10 23 19 32 47 63 79 96 91 97 Remove 32 Find successor and detach

temp

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Removed node has 2 children

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Replacement with successor

57 59 54 53 44 41 37 30 12 7 10 23 19 32 47 63 79 96 91 97 Remove 32 Find successor and detach

temp

37

temp

Attach removed node’s children to its successor

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Removed node has 2 children

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Replacement with successor

57 59 54 53 44 41 30 12 7 10 23 19 32 47 63 79 96 91 97 Remove 32 Find successor and detach Attach removed node’s children to its successor 37

temp

Make successor child of removed node’s parent

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Removed node has 2 children

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Replacement with successor

57 59 54 53 44 41 30 12 7 10 23 19 47 63 79 96 91 97 37 Remove 32 Find successor and detach Attach removed node’s children to its successor Make successor child of removed node’s parent In this example the successor had no subtree

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Removed node has 2 children

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Replacement with predecessor

57 59 54 53 44 41 37 30 12 7 10 23 19 32 47 63 79 96 91 97

temp

Remove 63 Find predecessor

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Removed node has 2 children

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Replacement with predecessor

57 59 54 53 44 41 37 30 12 7 10 23 19 32 47 63 79 96 91 97

temp

Remove 63 Find predecessor Attach predecessor’s subtree to its parent

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Removed node has 2 children

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Replacement with predecessor

57 59 54 53 44 41 37 30 12 7 10 23 19 32 47 63 79 96 91 97

temp

Remove 63 Find predecessor Attach predecessor’s subtree to its parent 59

temp

Attach removed node’s children to predecessor

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Removed node has 2 children

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Replacement with predecessor

57 54 53 44 41 37 30 12 7 10 23 19 32 47 63 79 96 91 97 Remove 63 Find predecessor Attach predecessor’s subtree to its parent 59

temp

Attach deleted node’s children to predecessor Attach predecessor to deleted node’s parent

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Removed node has 2 children

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Replacement with predecessor

57 54 53 44 41 37 30 12 7 10 23 19 32 47 79 96 91 97 Remove 63 Find predecessor Attach predecessor’s subtree to its parent 59

temp

Attach removed node’s children to predecessor Attach predecessor to removed node’s parent

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BST removal alternatives

• An alternative to the removal approach for nodes with 2

children is to replace the data

– The data from the predecessor node is copied into the node to be deleted – And the predecessor node is then removed

• Using the approach described for removing nodes with one or no children
• This avoids some of the complicated pointer assignments

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BST efficiency

• The efficiency of BST operations depends on the height of the

tree

– All three operations (search, insert and delete) are O(height)

• If the tree is complete the height is log(n)

– What if it isn’t complete?

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Height of a BST

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Insert 7 Insert 4 Insert 1 Insert 9 Insert 5 7 4 1 9 5 This is a complete BST height = log(5) = 2

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Height of a BST

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Insert 9 Insert 1 Insert 7 Insert 4 Insert 5 This is a linked list with extra unused pointers 9 1 7 4 5 height = n – 1 = 4 = O(n)

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Balanced BSTs

• It would be ideal if a BST was always close to complete

– i.e. balanced

• How do we guarantee a balanced BST?

– We have to make the structure and / or the insertion and removal algorithms more complex

• e.g. red – black trees, AVL trees (next!)

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Readings for this lesson

• Carrano & Henry

– Chapter 15.3 (Binary search tree) – Chapter 16.3 (Implementation – BST)

• Next class:

– Carrano & Henry: Chapter 19.5 (AVL tree)

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