Programming Languages and Techniques (CIS120)

Transcription

1 Programming Languages and Techniques (CIS120) Lecture 7 January 30 th, 2015 Binary Search Trees (Lecture notes Chapter 7)

2 ! Homework #1 feedback dope assignment, got a 96.4 the first time, careless mistake, fixed it. Hopefully this is a 100. Here we go I thought this was a good first assignment because it allowed me to get familiar with OCaml without being too difficult. Simple, but I had a really hard time setting up ocaml, I ended up finally setting it up on the 27th at 5 pm! I really liked the simplicity of this homework, but also how it let me become fairly familiar with Ocaml.! Pattern matching is cool :D Hard New to Ocaml, kind of get the hang of it I had trouble with the insertion sort tracery Let's do this! I thought it was cool. Some things that would be difficult in java seem really simple in OCaml so far. Homework got easier as I gained more understanding of Ocaml.

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7 Trees as containers Big idea: find things faster by searching less

8 Binary Search Trees (BST) Empty is a BST Node(lt,x,rt) is a BST if - lt and rt are both BSTs - all nodes of lt are x - all nodes of rt are > x

9 A Binary Search Tree 5 > 1 > > Node(lt,x,rt) is a BST if - lt and rt are both BSTs - all nodes of lt are x - all nodes of rt are > x Empty is a BST 8

10 Is this a BST?? 1. yes 2. no Node(lt,x,rt) is a BST if - lt and rt are both BSTs - all nodes of lt are x - all nodes of rt are > x Empty is a BST Answer: no, 7 to the lew of 6 7

11 Is this a BST?? 1. yes 2. no Node(lt,x,rt) is a BST if - lt and rt are both BSTs - all nodes of lt are x - all nodes of rt are > x Empty is a BST Answer: no, 5 to the lew of 4 8

12 Is this a BST?? 1. yes 2. no Node(lt,x,rt) is a BST if - lt and rt are both BSTs - all nodes of lt are x - all nodes of rt are > x Empty is a BST 8

13 Is this a BST?? 1. yes 2. no 4 Node(lt,x,rt) is a BST if - lt and rt are both BSTs - all nodes of lt are x - all nodes of rt are > x Empty is a BST Answer: yes

14 Is this a BST?? 1. yes 2. no Node(lt,x,rt) is a BST if - lt and rt are both BSTs - all nodes of lt are x - all nodes of rt are > x Empty is a BST

15 Is this a BST?? 1. yes 2. no Node(lt,x,rt) is a BST if - lt and rt are both BSTs - all nodes of lt are x - all nodes of rt are > x Empty is a BST Answer:Yes 6

16 Searching a BST (* Assumes that t is a BST *) let rec lookup (t:tree) (n:int) : bool = begin match t with Empty -> false Node(lt,x,rt) -> if x = n then true else if n x then (lookup lt n) else (lookup rt n) end The BST invariants guide the search. Note that lookup may return an incorrect answer if the input is not a BST!

17 Construc\ng BSTs

18 How do we construct a BST? Op\on 1: Build a tree Check that the BST invariants hold Op\on 2: Write func\ons for building BSTs from other BSTs e.g. insert an element, delete an element, Star\ng from some trivial BST (e.g. Empty), apply these func\ons to get the BST we want If each of these func\ons preserves the BST invariants, then any tree we get from them will be a BST by construc+on No need to check!

19 Inser\ng a new node: (insert t 4) > 1 4 > 1 > > > 3 9? 8

20 Inser\ng a new node: (insert t 4) 5 > 1 > > > 4 8

21 Inser\ng Into a BST (* Insert n into the BST t *) let rec insert (t:tree) (n:int) : tree = begin match t with Empty -> Node(Empty,n,Empty) Node(lt,x,rt) -> if x = n then t else if n x then Node(insert lt n, x, rt) else Node(lt, x, insert rt n) end Note the similarity to searching the tree. Note that the result is a new tree with one more Node; the original tree is unchanged Assuming that t is a BST, the result is also a BST. (Why?)

22 If you have seen BST insert before, was it wricen in an impera\ve or func\onal style? 1. Func\onal: inser\on returned a new tree that included the new element 2. Impera\ve: inser\on modified the tree to contain the new element 3. I don t remember 4. I m new to BSTs

23 Dele\on No Children: (delete t 3) > 1 3 > 1 > >

24 Dele\on No Children: (delete t 3) 5 > 1 7 > 0 9 If the node to be deleted has no children, simply replace it by the Empty tree. 8

25 Dele\on One Child: (delete t 7) 5 7 > 5 > 1 > >

26 Dele\on One Child: (delete t 7) 5 > If the node to be delete has one child, replace the deleted node by its child. 1 >

27 Dele\on Two Children: (delete t 5) 5 > 1 > >

28 Dele\on Two Children: (delete t 5) 3 > If the node to be delete has two children, promote the maximum child of the lew tree. 1 7 >

29 Would it also work to move the smallest label from the right- hand subtree? 1. yes 2. no Answer: yes

30 Subtle\es of the Two- Child Case Suppose Node(lt,x,rt) is to be deleted and lt and rt are both themselves nonempty trees. Then: 1. There exists a maximum element, m, of lt (Why?) 2. Every element of rt is greater than m (Why?) To promote m we replace the deleted node by: Node(delete lt m, m, rt) I.e. we recursively delete m from lt and relabel the root node m The resul\ng tree sa\sfies the BST invariants

31 How to Find the Maximum Element? What is the max element of this subtree?

32 How to Find the Maximum Element? Just for fun, how do we find the max element of the whole tree?

33 Tree Max: A par+al* func\on let rec tree_max (t:tree) : int = begin match t with Node(_,x,Empty) -> x Node(_,_,rt) -> tree_max rt _ -> failwith tree_max called on Empty end We never call tree_max on an empty tree This is a consequence of the BST invariants and the case analysis done by the delete func\on BST invariant guarantees that the maximum- value node is farthest to the right * Par\al, in this context, means not defined for all inputs.

34 Code for BST delete trees.ml

35 Dele\ng From a BST let rec delete (t: tree) (n: int) : tree = begin match t with Empty -> Empty Node(lt, x, rt) -> if x = n then begin match (lt, rt) with (Empty, Empty) -> Empty (Node _, Empty) -> lt (Empty, Node _) -> rt _ -> let m = tree_max lt in Node(delete lt m, m, rt) end else if n x then Node(delete lt n, x, rt) else Node(lt, x, delete rt n) end

36 If we insert a label n into a BST and then immediately delete n, do we always get back a tree of exactly the same shape? 1. yes 2. no Answer: no, what if the node is in the tree

37 If we insert a label n into a BST that does not already contain n and then immediately delete n, do we always get back a tree of exactly the same shape? 1. yes 2. no Answer: yes

38 If we delete n from a BST (containing n) and then immediately insert n again, do we always get back a tree of exactly the same shape? 1. yes 2. no Answer: no, what if we delete the root?