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BST Trees

5. 3. 8. 1. 4. 9. BST Trees. 5. 3. 8. 1. 4. 9. Binary search tree (BST). A binary search tree is a binary tree in which every node satisfies the following: the key of every node in the left subtree is smaller than the key of this node

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BST Trees

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  1. 5 3 8 1 4 9 BST Trees

  2. 5 3 8 1 4 9 Binary search tree (BST) • A binary search tree is a binary tree in which every node satisfies the following: • the key of every node in the left subtree is smaller than the key of this node • the key of every node in the right subtree is larger than the key of this node • Note: Duplication nodes ?

  3. 5 3 8 1 4 9 Binary search tree • The left subtree and the right subtree are also BST (Recursion) • No two nodes in a BST have the same key • If we traverse a BST inorder, we list the key in ascending order

  4. 8 1 4 12 9 2 10 14 6 7 1 9 11 3 7 15 5 13 10 3 2 12 5 6 14 3 8 16 4 15 Some examples 8

  5. This is an incorrect check of BST. 5 Every node satisfies the following: -- the key of the left node is smaller than the its own key -- the key of the right node is larger than its own key 3 8 1 7 9 Is this a BST? This is NOT BST!

  6. To find a key in a BST, we start at the root. Compare the target key with the key of the current node. - target == p->key, Done. - target < p->key, go left - target > p->key, go right 8 4 12 10 2 14 6 1 9 11 3 7 15 13 5 Note that the principle behind is similar to binary search... Find 5 Search BST target

  7. Basic Functions • Search • Find Min • Find Max • Insert(x) • Delete

  8. C++ Implementation of BST Search // Internal method to find an item in a subtree. // x is item to search for // t is the node that roots the tree // Return node containing the matched item. template <class Comparable> BinaryNode<Comparable> * BinarySearchTree<Comparable>:: find( const Comparable & x, Node *t ) const { while( t != NULL ) { if( x < t->element ) // element is the key value for each node t = t->left; else if( t->element < x ) t = t->right; else return t; // Match } return NULL; // Not found }

  9. 8 4 12 2 10 6 14 1 9 11 3 7 15 13 5 Balance tree The efficiency of BST search depends on the shape of the tree. If the tree is balance, (minimum height, very bushy), the search is fast (log(n) steps). If the tree is a long narrow chain, the search is slow (n steps) 1 15 2 14 Compare this with sequential and binary search ... 3 log(16)=4 4 5 13

  10. Binary Search Tree Class - FindMin/Max template <class Comparable> BinaryNode<Comparable> * BinarySearchTree<Comparable>::findMin( Node *t ) const{ if( t != NULL ){ while( t->left != NULL ) t = t->left; // the leftmost node } return t; } template <class Comparable> BinaryNode<Comparable> * BinarySearchTree<Comparable>::findMax( Node *t ) const{ If( t != NULL ) while( t->right != NULL ) t = t->right; // the rightmost node return t; }

  11. Insert & Delete in BST • First we’ll study simple-minded Insert and Delete procedure. They may result in highly unbalanced tree • Then we’ll study a special kind of BST called AVL tree which maintains near balance in inserting and deleting.

  12. 10 5 12 4 7 14 15 2 1 9 3 17 8 Insert in BST (recursive version) • // Internal method to insert into a subtree • // x is the item to insert • // t is the node that roots the tree • // Set the new root • // Throw an exception if x is already in t • template <class Comparable> • void BinarySearchTree<Comparable>:: • insert( const Comparable & x, Node * & t ) const { • if( t == NULL ) • t = new Node( x, NULL, NULL ); • else if( x < t->element ) • insert( x, t->left ); • else if( x > t->element) • insert( x, t->right ); • else • throw DuplicateItemException( ); • } 6 16 New nodes are always added as leaves.

  13. Delete in BST Case 1: If it has no children, that is, it is a leaf, we just remove it.

  14. L  L Delete in BST Case 2: If it has only one child, (i.e. only one subtree), we splice out it. That is, we attach that subtree to the parent of the node.

  15. 10 10 10 7 5 12 12 7 12 7 14 14 6 9 6 9 17 6 9 17 17 15 15 15 16 16 16 Delete in BST, examples   Case 2: delete a node that has only one subtree.

  16. Min in R  L R’ L R Delete in BST • Case 3: If it has nonempty left and right subtrees. • Find the minimum node in right subtree, then copy its key value to X, and remove the min node from right subtree. • Or Find the min node in right subtree, then swap it with the target node, and remove the target node

  17. 4 2 3 1 Delete Tree, example 10 10  5 12 6 12 4 8 8 16 16 2 6 9 7 9 14 14 3 7 1 15 15 Case 3: delete a node that has nonempty left and right subtrees.

  18. Delete in BST template <class Comparable> void BinarySearchTree<Comparable>:: remove( const Comparable & x, Node * & t ) const { if( t == NULL ) throw ItemNotFoundException( ); if( x < t->element ) remove( x, t->left ); else if( t->element < x ) remove( x, t->right ); else if( t->left != NULL && t->right != NULL ) { //case 3 t->element = findMin( t->right )->element; removeMin( t->right ); // Remove minimum } Else { // case 1, 2 BinaryNode<Comparable> *oldNode = t; t = ( t->left != NULL ) ? t->left : t->right; // Reroot t delete oldNode; // delete old root } }

  19. Order statistics • findKth smallest element problem based on the consider of relationship of node size and K. X SL SL SR SR SR X X SL If K == SL +1, it is the root If K > SL +1, it is the K-(SL +1)th smallest on the right subtree If K < SL +1, it is the Kth samllest element in the left subtree Note: SL is the size of the left subtree

  20. Binary Search Tree • Binary search tree, using simple insert and delete procedures • the tree is nearly balance • add - fastO(log n) • delete a target - fastO(log n) • search - fastO(log n) • the tree is highly unbalance, it becomes a singly linked list (the worst case) • add, delete and search - slowO(n)effectively using sequential search to find a location

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