CS 151. Binary Search Trees. Wednesday, October 10, 12

Transcription

1 CS 151 Binary Search Trees 1

2 Binary Search Tree A binary search tree stores comparable elements in a binary tree such that for every node r all nodes in the left BST are <= element at r all nodes in the right BST are > element at r Ex. 2 2

3 BST Operations A binary search tree stores comparable elements at its nodes. Operations are: standard tree operations such as size(), isempty(), leafcount(), etc. insert(x) - insert element x into its correct position in the tree remove(x) - remove x from the tree, if it exists removemin() - remove the minimum element from the tree (where is it?) removemax() - remove the maximum element from the tree (where is it?) find(x) - return item that matches x findmin() - return the minimum element of the tree findmax() - return the maximum element of the tree traverse()/iterate() - go through the nodes of the tree in increasing order 3 3

4 BST Implementation Although you can implement a BST as we did with the binary tree, we ll try something different: } // BST constructor public BST( T data, BST<T> left, BST<T> right ) this.data = data this.left = left this.right = right // what to initialize size to? // can you construct an empty tree with this implementation? 4 4

5 BST Implementation Although you can implement a BST recursively, we ll try non-recursively: public BST( T data, BST<T> left, BST<T> right ) this.data = data this.left = left this.right = right if( data == null ) size = 0 // empty trees have null data else size = 1 if( left!= null ) size += left.size() if( right!= null ) size += right.size() } public BST( ) // construct an empty tree this( null, null, null ) 5 5

6 BST Implementation First let s try implementing some of the standard binary tree methods: public boolean isempty() return size == 0 // or, (data == null) public boolean isleaf() if isempty() return false return (left == null) && (right == null) // or, (size==1) public int leafcount() if isempty() return 0 if isleaf() return 1 if( left!= null ) count += left.leafcount() if( right!= null ) count += right.leafcount() return count 6 6

7 A couple more: height and inorder BST Implementation public int height() if isempty() return -1 if isleaf() return 0 if( left!= null ) lefth = left.height() if( right!= null ) righth = right.height() return 1 + max{ lefth, righh} public void inorder( List<T> L ) if isempty() return if( left!= null ) left.inorder( L ) L.add( data ) if( right!= null) left.inorder( L ) 7 7

8 BST Implementation Finally, let s try breadth-first search (bfs), aka level-order traversal: public void bfs( List<T> L ) if( isempty() ) return Queue<T> queue = new Queue<T>() queue.add( this ) // add this root node to the queue while(!queue.isempty() ) // while queue is not empty front = queue.front() // get the front node L.add( front.getdata() )// enumerate the front node if( left!= null ) // if L child exists queue.enqueue( left ) // add it to queue if( right!= null ) // if R child exists queue.enqueue( right ) // add it to queue 8 8

9 Now, let s talk about the new BST methods. First up: find(x) BST Implementation public BST<T> find(t x) // return node containing x if( isempty() ) return null // didn t find it if( x.compareto( data ) < 0 ) // x < data if( left == null ) return (left = new BST(x,null,null)) return left.find( x ) // search in the left subtree else if( x.compareto( data ) > 0 ) // x > data if( right == null ) return (right = new BST(x,null,null)) return right.find( x ) // search in the right subtree else return data // x == data, found it! } // Question: what is the running time of find? // Answer: O(height of the tree) = O( n ) 9 9

10 Let s try insert(x) -- always add new node as a leaf node 10 BST Implementation public void insert(t x) if( x == null ) throw Exception // don t allow null adds if( isempty() ) data = x // make a single node else if( x.compareto( data ) < 0 )// if x<data, insert left if( left == null ) left = new BST( x, null, null ) else left.insert( x ) else if( x.compareto( data ) > 0 )// if x>data, insert right if( right == null ) right = new BST( x, null, null) else right.insert( x ) else throw DuplicateException // x == data, no duplicates size++ } // Question: What is the running time of insert? O(height)=O(n) 10

11 Now for findmax(x). findmin(x) is very similar BST Implementation public T findmax(t x) if( isempty() ) return null / throw NoSuchElementException? if( right == null ) // if no right child, this is the max return data return right.findmax( x ) } //Question: what is the running time of findmax? O(height)=O(n) 11 11