1 Trees 3: The Binary Search Tree Section 4.3

2 Binary Search Tree A binary tree B is called a binary search tree iff: –There is an order relation < defined for the vertices of B –For any vertex v, and any descendant u in the subtree v.left, u < v –For any vertex v, and any descendent w in the subtree v.right, v < w root

3 Binary Search Tree Which one is NOT a BST?

4 Binary Search Tree Consequences –The smallest element in a binary search tree (BST) is the “left-most” node –The largest element in a BST is the “right-most” node –Inorder traversal of a BST encounters nodes in increasing order root

5 Binary Search using BST Assumes nodes are organized in a binary search tree –Begin at root node –Descend using comparison to make left/right decision if (search_value < node_value) go to the left child else if (search_value > node_value) go to the right child else return true (success) –Until descending move is impossible –Return false (failure)

6 Binary Search using BST Runtime <= descending path length <= depth of tree If tree has “enough” branching, runtime is O(log n) –Worst case is O(n)

7 BST Class Template

8 BST Class Template (contd.) Internal functions used in recursive calls Pointer passed by reference (why?)

9 BST: Public members calling private recursive functions

10 BST: Searching for an element

11 BST: Find the smallest element Tail recursion

12 BST: Find the biggest element Non-recursive

13 BST: Insertion (5) Before insertion After insertion

14 BST: Insertion (contd.) Strategy: Traverse the tree as in searching for t with contains() Insert if you cannot find t

15 BST: Deletion of Leaf Before deleting (3) After deleting (3) Deleting a node with no child Deletion Strategy: Delete the node

16 BST: Delete a Node with One Child Before deleting (4) After deleting (4) Deleting a node with one child Deletion Strategy: Bypass the node being deleted

17 BST: Delete a Node with Two Children Before deleting (2) After deleting (2) Deleting a node with two children Replace the node with smallest node in the right subtree

18 BST: Deletion Code

19 BST Deletion Element: 5 Left: 208 Right: 0 Element: 3 Left: 0 Right: 160 Element: 4 Left: 0 Right: Address 208 Address 160 Element: 5 Left: 160 Right: 0

20 BST: Lazy Deletion Another deletion strategy –Don’t delete! –Just mark the node as deleted –Wastes space –But useful if deletions are rare or space is not a concern

21 BST: Insertion Bias Start with an empty tree Insert elements in sorted order What tree do you get? How do you fix it?

22 BST: Deletion Bias After large number of alternating insertions and deletions Why this bias? How do you fix it?

23 BST: Search using function objects

24 Average Search/Insert Time - 1 Average time is the average depth of a vertex –Let us compute the sum of the depths of all vertices and divide by the number of vertices –The sum of the depths is called the internal path length Give the internal path lengths for the following trees

25 Average Search/Insert Time Let D(N) be the internal path length of a tree with N vertices If the root has a left subtree with i nodes, then D(N) = D(i) + D(N-i-1) + N-1 because the depth of each vertex in the subtrees increases by

26 Average Search/Insert Time - 3 Root Subtree with N-1 nodes The average value of D(N) is given by the recurrence D(1) = 0 D(N) = 1/N[  i=0 N-1 D(i) + D(N-i-1)] + N - 1 = 2/N  i=0 N-1 D(i) + N - 1 Root Subtree with N-2 nodes Subtree with 1 node Root Subtree with N-3 nodes Subtree with 2 nodes

27 Average Search/Insert Time - 4 D(N) = 2/N  i=0 N-1 D(i) + N - 1 N D(N) = 2  i=0 N-1 D(i) + N(N - 1) (1) (N-1)D(N-1) = 2  i=0 N-2 D(i) + (N-1)(N - 2) (2) (2) - (1) gives ND(N) - (N-1)D(N-1) = 2D(N-1) + 2(N-1) ND(N) = (N+1)D(N-1) + 2(N-1) D(N)/(N+1) = D(N-1)/N + 2(N-1)/[N(N+1)] < D(N-1)/N + 2/N D(N)/(N+1) < D(N-1)/N + 2/N D(N-1)/(N) < D(N-2)/(N-1) + 2/(N-1) D(N-2)/(N-1) < D(N-3)/(N-2) + 2/(N-2)... D(2)/(3) < D(1)/2 + 2/2

28 Average Search/Insert Time - 5 D(N)/(N+1) < D(N-1)/N + 2/N < D(N-2)/(N-1) + 2/(N-1) + 2/N < D(N-3)/(N-2) + 2/(N-2) + 2/(N-1) + 2/N... < D(1)/(2) + 2/ /(N-2) + 2/(N-1) + 2/N = 2  i=2 N 1/i If we show that  i=2 N 1/i is O(log N), then we can prove that average D(N) = O(N Log N) and so the average depth is O(log N) D(N)/(N+1) < D(N-1)/N + 2/N D(N-1)/(N) < D(N-2)/(N-1) + 2/(N-1) D(N-2)/(N-1) < D(N-3)/(N-2) + 2/(N-2)... D(2)/(3) < D(1)/2 + 2/2

29 Deriving Time Complexity Using Integration Integration can be used to derive good bounds for sums of the form  i=a N f(i) when f(i) is monotonically increasing or decreasing if you know how to integrate f(x) / 2 f(x) = 1/x 1/ 3 1/ 4 Area under the rectangles is smaller than that under 1/x  i=2 4 1/i < ∫ 1 4 1/x dx  i=2 N 1/i < ∫ 1 N 1/x dx = ln (N) - ln (1) = O(log N)