Binary Search Trees 15-121 Fall 2020 Margaret Reid-Miller Today - - PowerPoint PPT Presentation

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Binary Search Trees 15-121 Fall 2020 Margaret Reid-Miller Today - - PowerPoint PPT Presentation

Nov 17, 2023 206 likes •566 views

Binary Search Trees 15-121 Fall 2020 Margaret Reid-Miller Today Prune leaves example Binary Search Trees Definition Algorithms for Contains Add Remove Complexity Fall 2020 15-121 (Reid-Miller) 2 Remove all

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Binary Search Trees

15-121 Fall 2020 Margaret Reid-Miller

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Today

  • Prune leaves example
  • Binary Search Trees
  • Definition
  • Algorithms for
  • Contains
  • Add
  • Remove
  • Complexity

Fall 2020 15-121 (Reid-Miller) 2

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Remove all leaves from tree t

public static <E> void prune(BTNode<E> t){ if (t == null) return; BTNode<E> left = t.getLeft(); if (left != null) { if (isLeaf(left)) t.setLeft(null); else } BTNode<E> right = t.getRight(); if (right != null) { if (isLeaf(right)) t.setRight(null); else } }

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prune(left); prune(right);

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Remove all leaves from tree t Return reference to the resulting tree

public static <E> BTNode<E> prune(BTNode<E> t){ What are the base case(s)? What does the recursive cases return? How do you update the tree? Submit answers to

https://forms.gle/a5qj8rTF1tndp1tH6

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Remove all leaves from tree t Return reference to the resulting tree

public static <E> BTNode<E> prune(BTNode<E> t){ if (t == null || isLeaf(t)) return null; t.setLeft = prune(t.getLeft()); t.setRight = prune(t.getRight()); return t; }

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More Terminology

  • A perfect binary tree is a binary tree such that
  • all leaves have the same level, and
  • every non-leaf node has 2 children.
  • A full tree is a binary tree such that
  • every non-leaf node has 2 children.
  • A complete binary tree is a binary tree such that
  • every level of the tree has the maximum

number of nodes

  • except possibly at the deepest level, where the

nodes are as far left as possible.

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Examples

A G F E D C B F A E D C B

PERFECT (COMPLETE and FULL) COMPLETE (but not PERFECT and not FULL)

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Binary Search Trees

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Binary Search Trees

  • A binary tree T is a binary search tree if
  • T is empty, or
  • T has two subtrees, TL and TR, such that
  • All the values in TL are less than the value

in the root of T,

  • All the values in TR are greater than the

value in the root of T, and

  • TL and TR are binary search trees
  • Typically, duplicates are not allowed.

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Example: Is this a BST?

10

4 7 1 6 9 8 3

yes no

3 4 7 1 6 9 8 3

Fall 2020 15-121 (Reid-Miller)

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Elements of a BinarySearchTree must be Comparable

public class BinarySearchTree <E extends Comparable<E>> { private class BTNode { // inner class private E data private BTNode left; private BTNode right; private BTNode(E data) { this.data = data} } private BTNode root; // field public BinarySearchTree() { // constructor root = null; }

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Example : max

How did we find the maximum value in a Binary Tree? Recursively got the max of the root and the left and right subtrees. If the tree is a Binary Search Tree (BST), do we need to search the whole tree? No Where is the maximum in a Binary Search Tree? The right most node

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Where is the maximum? 15 40 26 45 62 11 96 39 83 21 52 45

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96

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The node with the maximum value in a BST is the rightmost node.

// recursive implementation public E max() { if (root == null) return null; return max(root); } //precondition: tree is not empty private E max(BTNode t) { if (_______________) return t.data; else return max(t.right); }

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Only one recursive call!

t.right == null

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Searching for 83 and 42 62 11 96 39 83 21 45 62 11 96 39 83 21 45 62 96 83 62 11 39 45

NOT FOUND

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// iterative implementation public boolean contains(E obj) { BTNode t = root; while (t != null) { if (obj.equals(t.data)) return true; else if (obj.compareTo(t.data) < 0) t = t.left; else t = t.right; } return false; }

To find an element follow a path in the BST

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Adding to a BST

  • Where should we add an element to a BST?
  • If the tree is empty,

make the element the root

  • Otherwise attach it as a leaf node.

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Add to a BST in the order given

62 96 11 39 21 83 45

62 11 96 39 83 21 45

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Add element as a leaf node

// recursive implementation public void add(E obj) { root = add(root, obj); } // returns a reference to a tree with obj in it private BTNode add(BTNode t, E obj) { if (t == null) return new BTNode(obj); else if (obj.compareTo(t.data) < 0) t.left = add(t.left, obj); else t.right = add(t.right, obj); // else already in the tree return t; }

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if (obj.compareTo(t.data) > 0)

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Exercise: add

  • Draw the binary search tree that results from

adding the each character to an empty tree in the following order: C A R N E G I E M E L L O N Do not add any duplicate values to your tree.

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C A R N E G I E M E L L O N

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Runtime of add on a perfect BST

  • A perfect binary search tree with height H has how

many nodes?

  • n = 1 + 2 + 4 + ... + 2H = 2H+1 - 1
  • Thus, H = log2(n+1) - 1.
  • Add will take O(log n) time on a perfect BST
  • We have to examine one node at each level before

we find the insertion point, and height is H.

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Worst-case runtime of add on a BST

  • Are all BSTs perfect?

Add the following numbers in order into a BST: 11 21 39 45 62 83 96 What do you get?

  • An add will take O(n) time in the worst case on an

arbitrary BST since there may be up to n levels in a BST with n nodes.

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NO!

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Aside: How many BSTs with n distinct values have height n?

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1 2 3 4 5 5 4 3 2 1 1 4 5 3 2

Total number BST with height n is 2n-1 That's a small fraction of the total number BST: O(4n)

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Removing from a BST

General idea:

  • Search for the item to remove until either:
  • the item is found (at the root of a subtree) or
  • we reach a leaf and the item is not found
  • If the item is found:
  • remove the root (of the subtree) and
  • replace it with its closest value

(either its in-order predecessor or successor)

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K D R O W C E L H S N M A T

Predecessor is largest value in left subtree Successor is smallest value in right subtree

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Suppose we replace the root with its in-order predecessor.

Where is the in-order predecessor?

  • n the left side and
  • is the rightmost node (max on left side)

Three cases:

  • 1. no predecessor
  • 2. predecessor is left child
  • 3. predecessor is rightmost of left child

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Case 1: The root has no predecessor

  • Return the right subtree (to the parent of the

root)

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r

S S

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Removing from a BST

Data value is found at a root with no left child

  • Return the right subtree
  • Result is to set the parent reference to its grandchild.
  • Example:

Remove 31. null

64

null 31

49

X

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root:

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Case 2: The predecessor is the left child of the root

  • Copy the data from the root's left child into the root.
  • Set the root's left reference to the left child of the root's

left child.

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r p

S A

p

A S

root root Subtree A could be null

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Removing from a BST

Predecessor is left child

  • Example:

Remove 29. 45 29 17 null 36 6 17

X

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predecessor of 29 root

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Case 3: The root’s predecessor is rightmost of the left child

  • Copy the data from the predecessor to the root
  • Then set the predecessor's parent to reference its left

child.

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r

A S G

p p

A S G

root root Subtree G could be null

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Removing from a BST

Predecessor is right of left child

  • Example:

Remove 29. 45 29 15 36 6 21 27 null 27

predecessor of 29

24

X

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root

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Complexity of Binary Search Trees

Worst case Best case contains O(height) O(1) add O(height) O(1) remove O(height) O(1) traverse O(n) Worst-case height of binary search tree: O(n) Best-case height of binary search tree: O(log n) Balanced Binary Search Trees have height O(log n): E.g, AVL, Red-Black, Splay, Treaps, Weight-balanced, …

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