Binary Search Trees CS16: Introduction to Data Structures & - PowerPoint PPT Presentation

Binary Search Trees CS16: Introduction to Data Structures & Algorithms Spring 2020

Outline ‣ Binary Search Trees ‣ Searching BSTs ‣ Adding to BSTs ‣ Removing from BSTs by CynthT http://cyntht.deviantart.com/ ‣ BST Analysis ‣ Balancing BSTs

Binary Search Trees ‣ Binary trees with special property ‣ For each node ‣ left descendants have lower value than node ‣ right descendants have higher value than node ‣ In-order traversal gives nodes in order 3

Searching a BST 13 11 < 13 7 20 11 > 7 5 10 15 24 11 > 10 8 11 ‣ Find 11 ‣ Each comparison tells us whether to go left or right 4

Searching a BST 13 14 > 13 7 20 14 < 20 5 10 15 24 reached leaf w/o finding 8 11 it so not in tree ‣ What if item isn’t in tree? ‣ Find 14 5

Binary Search Tree — Find( ) function find (node, toFind): if node.data == toFind: return node else if toFind < node.data and node.left != null: return find(node.left, toFind) else if toFind > node.data and node.right != null: return find(node.right, toFind) return null 6

Inserting in a BST 13 17 > 13 7 20 17 < 20 5 10 15 24 17 > 15 17 8 11 ‣ To insert, perform a search and add as new leaf ‣ Insert 17 7

Binary Search Tree — Insert( ) function insert (node, toInsert): if node.data == toInsert: # data already in tree return if toInsert < node.data: if node.left == null: # add as left child node.addLeft(toInsert) else: insert(node.left, toInsert) else: if node.right == null: # add as right child node.addRight(toInsert) else: insert(node.right, toInsert) 8

Removing from a BST Can be tricky ‣ 13 Three cases to consider ‣ ‣ Removing a leaf: easy, just do it 7 20 ‣ Removing internal node w/ 1 child (e.g., 15 ) 5 10 15 24 ‣ Removing internal node w/ 2 children (e.g., 7 ) 17 8 11

Removing from a BST - Case #2 Removing internal node w/ 1 child ‣ 13 Strategy ‣ ‣ “Splice out” node by connecting its 7 20 parent to its child Example: remove 15 ‣ 5 10 15 24 set parent’s left pointer to 17 ‣ remove 15’s pointer ‣ 17 8 11 no more references to 15 so erased ‣ (garbage collected) BST order is maintained ‣

Removing from a BST - Case #3 Removing internal node w/ 2 children ‣ 13 Replace node w/ successor ‣ successor: next largest node ‣ 7 20 ‣ Delete successor 5 10 17 24 ‣ Successor a.k.a. the in-order successor 8 11 ‣ Example: remove 7 ‣ What is successor of 7 ? 9

Removing from a BST - Case #3 Since node has 2 children… ‣ 13 …it has a right subtree ‣ Successor is leftmost node ‣ 7 20 in right subtree 5 10 17 24 ‣ 7 ’s successor is 8 successor (node): 8 11 curr = node.right while (curr.left != null): curr = curr.left 9 return curr

Removing from a BST - Case #3 Now, replace node with successor ‣ 13 Observation ‣ Successor can’t have left sub-tree ‣ 7 8 20 …otherwise its left child ‣ would be successor 5 10 17 24 so successor only has right child ‣ Remove successor using ‣ 8 11 Case # 1 or # 2 Here, use case #2 (internal w/ 1 child) ‣ 9 Successor removed and BST order restored ‣

Binary Search Tree — Remove( ) function remove (node): if node has no children: # case 1 node.parent.removeChild(node) else if node only has left child: # case 2a if node.parent.left == node: # node is a left child node.parent.left = node.left else: node.parent.right = node.left else if node only has right child: # case 2b if node.parent.left == node: node.parent.left = node.right else: node.parent.right = node.right else: # case 3 (node has two children) nextNode = successor(node) node.data = nextNode.data #replace w/ nextNode remove(nextNode) # nextNode has at most one child 14

Successor vs. Predecessor ‣ In Remove( ) ‣ OK to remove in-order predecessor instead of in-order successor ‣ Randomly picking between the two keeps tree balanced ‣ In Case # 3 ‣ Predecessor is rightmost node of left subtree 15

Binary Search Tree — Remove( ) 2 min Activity #1 16

Binary Search Tree — Remove( ) 2 min Activity #1 17

Binary Search Tree — Remove( ) 1 min Activity #1 18

Binary Search Tree — Remove( ) 0 min Activity #1 19

Binary Search Tree Analysis 13 ‣ How fast are BST operations? 7 20 ‣ Given a tree, what is the worst- case node to find/remove? 5 10 15 24 ‣ What is the best-case tree? D ‣ a balanced tree ‣ What is the worst-case tree? C ‣ a completely unbalanced tree B A

Binary Search Trees — Rotations ‣ We can re-balance unbalanced trees w/ tree rotations C B A A A B C C B ‣ In-order traversal of all 3 trees is Beyond CS16, But good to B A C know ‣ so BST order is preserved 21