CPS 100, Spring 2004, Written Tree Assignment
These problems provide practice with trees, recursion, and big-Oh.
You can talk about the problems with up other people, but each person
should submit his/her own version of the solutions. This means
your writeups should be your own work, even if you talk with
other people. It's in your own interest to do your own work
as test questions will be difficult if you cannot do these questions.
You may not use the Internet to do these problems, no Googling.
You should submit a file whose name must be tree.cpp, and a
README file that indicates how long you spent on these problems and with
whom you consulted.
You are to submit solutions to these problems using
submit_cps100 trees tree.cpp README
The solutions are in the file tree.cpp so that you can have your code
formatted automatically (e.g., if you use emacs/Eclipse). You should
NOT make your solutions run, you're just typing solutions to make them
easier to read and to allow for electronic submission. It is
important to think about your answers, not to run them.
Assume the following declarations have been made, note
that a TreeNode has a parent pointer, assume
this is properly initialized/set to point to the node's parent
in tree.
struct TreeNode
{
string info;
TreeNode * left;
TreeNode * right;
TreeNode * parent;
TreeNode (int val, TreeNode * lt, TreeNode * rt, TreeNode * p)
: info(val), left(lt), right(rt), parent(p)
{ }
};
- Binary Search Tree Iterator
- Write functions that return the minimum and
maximum nodes in a binary search tree rooted at
t. Assume trees store unique elements (no duplicates).
Provide a big-Oh expression for the complexity of the functions you
write, assuming trees are roughly balanced (depth = log(n) for n nodes).
TreeNode* minimum(TreeNode * t)
// pre: t is a search tree
// post: returns pointer to node with minimum key (NULL/0 if t is empty)
TreeNode* maximum(TreeNode * t)
// pre: t is a search tree
// post: returns pointer to node with maximum key (NULL/0 if t is empty)
- Write a function to find the inorder successor of a
node. You should use the parent link and the minimum function
(assume minimum works as specified). This function can be written
in fewer than 10 lines of code --- that's a guideline, not a requirement.
TreeNode* successor(TreeNode * t)
// pre: t is a search tree
// post: returns next node in inorder traversal of tree, returns NULL/0 if
// t is has no inorder successor (i.e., is maximal element)
- The following code
results in an inorder traversal of a tree from the root
void inOrder(TreeNode *root)
{
for (TreeNode *t = minimum(root); t != 0; t = successor(t)){
cout << t->info << endl;
}
}
What is the big-Oh of inOrder
in the worst case? Justify your answer briefly
-
Write the function levelCount whose header is given below.
levelCount returns the number of nodes on the specified
level.
For this problem, the root is at level zero, the root's children are
at level one, and for any node, the node's level is one more than its
parent's level. For example, for the bean-tree diagrammed below, the
call levelCount(t,1) should return 2 (chickpea and navy are
on level 1); the call levelCount(t,2) should return 3; and
the call levelCount(t,4) should return 0.
Hint: when the level is 0, there is always just one node at that level,
the root node (assuming t isn't empty). If the level isn't zero,
recursive calls will be used to determine the number of nodes at the
requested level.
int levelCount(TreeNode * t, int level)
// pre: 0 <= level
// post: returns # nodes at specified level in t
-
In the next two problems assume trees store int values rather
than string values
-
Write the function findKthInOrder whose header is given below.
Given an integer k and a binary search tree with unique values,
findKthInOrder returns a pointer to the node that contains the
kth element if the elements are in sorted order --- the smallest value
is k == 1, the second smallest is k == 2, and so on.
For example, in the tree t shown below (t points to the root),
findKthInOrder(t,4) returns a pointer to the node with value 9,
findKthInOrder(t,8) returns a pointer to the node with value 18,
and findKthInOrder(t,12) returns NULL/0 since there are only 9 nodes
in the tree.
You may find it useful to call the function count discussed
in lecture that returns the number of nodes in a tree.
int count(TreeNode * t)
// post: returns # nodes in t
{
if (t == 0) return 0;
return 1 + count(t->left) + count(t->right);
}
TreeNode * findKthInOrder(TreeNode * t, int k)
// pre: t is a binary search tree with unique values
// post: returns pointer to the kth node in sorted order,
// returns NULL if t has fewer than k nodes.
-
Assuming trees are roughly balanced (e.g., average case), write a
recurrence for your solution to findKthInOrder and give the
solution to the recurrence.
- Write a recurrence for your solution to findKthInOrder
for the worst case, i.e., when a tree is completely unbalanced
and the function is called to do the most work on every recursive call.
What is the solution to this recurrence?
- By definition, the
diameter of a tree (sometimes called the width) is the number
of nodes on the longest path between two leaves in the tree. The
diagram below shows two trees each with diameter nine, the leaves that
form the ends of a longest path are shaded (note that there is more
than one path in each tree of length nine, but no path longer than
nine nodes).
It can be shown that the diameter of a tree T is the largest
of the following quantities:
- the diameter of T's left subtree
- the diameter of T's right subtree
- the longest path between leaves that goes through the root of T
(this can be computed from the heights of the subtrees of T)
Here's code that's almost a direct translation of the three properties
above (assuming the existence of a standard O(1) max function that
returns the larger of two values).
int diameter(TreeNode * t)
// post: returns diameter of t
{
if (t == 0) return 0;
int leftD = diameter(t->left);
int rightD = diameter(t->right);
int rootD = height(t->left) + height(t->right) + 1;
return max(rootD, max(leftD, rightD));
}
-
However, the function as written does not run in O(n) time.
Write a recurrence for this implementation and what the solution
to the recurrence is. Assume trees are roughly balanced in writing
the recurrence.
- Write a version of diameter that runs in O(n) time.
Your function should use an auxiliary/helper function as described
below. The helper function should make
two recursive calls and do O(1) other work.
int diameterHelper (TreeNode * t, int & height)
// pre: t is a binary tree
// post: return (via reference param) height = height of t
// return as value of function: diameter of t
{
// write this
}
int diameter(TreeNode * t)
// post: returns diameter of t
{
int height;
return diameterHelper(t, height);
}
-
Two binary trees s and t are isomorphic if
they have the same shape; the values stored in the nodes do not affect
whether two trees are isomorphic. In the diagram below, the tree in
the middle is not isomorphic to the other trees, but the tree
on the right is isomorphic to the tree on the left.
Write a function isIsomorphic that returns true if its two
tree parameters are isomorphic and false otherwise. You must also
give the big-Oh running time (in the average case, assuming each tree
is roughly balanced) of your function with a justification.
Express
the running time in terms of the number of nodes in trees s
and t combined, i.e., assume there are n nodes together
in s and t with half the nodes in each tree.
bool isIsomorphic(TreeNode * s, TreeNode * t)
// post: returns true if s is isomorphic to t, else returns false
-
Two trees s and t are quasi-isomorphic if
s can be transformed into t by swapping left and
right children of some of the nodes of s. The values in the
nodes are not important in determining quasi-isomorphism, only the
shape is important. The trees below are quasi-isomorphic because if
the children of the nodes A, B, and G in the tree on the left are
swapped, the tree on the right is obtained.
Write a function isQuasiIsomorphic that returns true if two
trees are quasi-isomorphic. You must also give the big-Oh running
time (in the average case, assuming each tree is roughly balanced) of
your function with a justification. Express the running time in terms
of the number of nodes in trees s and t combined as
with the previous problem.
bool isQuasiIsomorphic(TreeNode * s, TreeNode * t)
// post: returns true if s is quasi-isomorphic to t, else returns false
Owen L. Astrachan
Last modified: Fri Feb 27 10:17:49 EST 2004