Question

For the set of keys [37, 24, 29, 66, 17, 82, 43], draw binary search trees of height 2, 4, and 6.

Argue that since sorting n elements takes Ω(n log n) time in the worst case in the comparison model, any comparison‐based algorithm for constructing a binary search tree from an arbitrary list of n elements takes Ω(n log n) time in the worst case.

When node z in TREE‐DELETE has two children, we could choose node y as its predecessor rather than its successor. What other changes to TREE‐DELETE would be necessary if we did so?

Answer 1

Detailed Explanation For sub-parts are given in images below:

Pseudo-code for Deletion is given after images...

2. Explanation for (n.logn) worst case for BST construction based on sorting.

Tree-Deletion Example and Explanation:

Pseudo-Code for Deletion Where Y as Predecessor:

Node Delete(Node Z):

Node Y = Z.right();

Node X = Z.left();

X = Insert(X.val, Y);

Z = Y;

Return Z;

Here Insertion() returns the pointer where the val is inserted..

Delete() returns the new pointer where deletion happened.