CodeSport (May 2009)

Welcome to another installment of ‘CodeSport’. This month we take a quick look at the problem of finding out whether a given binary tree is in fact a binary search tree. We then discuss the problem of finding the maximum and minimum in a binary search tree.

Thanks to all the readers who sent in their comments to the problems we discussed in last month’s column. Congratulations to Rohit Agarwalla and Rajeev Kumar for getting the solutions to the problem correct. Last month’s takeaway problem was on identifying binary search trees. A binary search tree (BST), or ordered binary tree, is a type of binary tree where the nodes are arranged in order: for each node in the tree, all elements in the left sub-tree of the node are less than or equal to the value of the node, and all the elements in the right sub-tree of the node are greater than the node. Binary search trees are used for element insertion and lookup.

Last month’s problem was to determine whether a given binary tree is, in fact, a binary search tree or not. Given a plain binary tree, you have to examine the tree to determine if it meets the requirements of a binary search tree, i.e., for every node, all of the nodes in its left tree must be less than or equal to the node, and all the nodes in its right subtree must be greater than the node. The algorithm should return true if the binary tree meets the criteria for a binary search tree. Else it should return false.

What is the best possible time complexity of such an algorithm? In order to verify that every node in the tree satisfies the binary search tree property, we need to visit each node once. Hence the best possible time complexity can not be less than O(N) where N is the number of nodes in the binary tree. Now how do we come up with an algorithm that verifies whether a binary tree satisfies the binary search tree property at each node?

We use the BST property that a left child of a node is less than or equal to the node, whereas the right child of a node is greater than the node. If we visit the left child first, then the node and then the right child, then these nodes will be visited in order. Such a walk on the nodes of the tree is known as ‘in order tree walk’. If we print the nodes as we visit each node, such a visit will print the nodes in monotonically increasing order. Hence all we need to do to verify that a binary tree is a BST, is to visit the nodes of the tree in order and print the elements. If such a walk results in a monotonically increasing sequence of elements, then the tree is a BST.

We can easily write a recursive algorithm to visit the binary tree in order. The steps of our recursive algorithm at each node x are as follows:

  1. Make sure that node x is not NULL
  2. Recursively print the elements in the left subtree of x
  3. Print the element at node x
  4. Recursively print the elements in the right subtree of x

We assume that each node x contains a pointer to its left child in the field ‘left’, a pointer to its right child in the field ‘right’, and the field ‘key’ gives the element value contained in the node x. We can assume the utility functions left(x), which returns the left child, right(x) which gives the right child, and key(x) which gives the value of the element at node x.

inOrderTreeWalk(x)
{
     if ( x is not null)
     {
          inOrderTreeWalk(left(x));
          print the value of key(x);
          inOrderTreeWalk(right(x));
     }
}

Given below is the pseudo code for the function IsThisTreeBST, which takes the root of the given binary tree as a parameter.

bool IsThisTreeBST(x)
{
     get S the sequence of elements returned by InorderTreeWalk(x);

     if (S is a monotonically increasing sequence)
          return true;
     else
          return false;
}

I leave it to the reader to write the code to verify that the sequence of elements returned by the inOrderTreeWalk is a monotonically increasing sequence or not.

This month’s programming question

In this month’s column, we will stay with the binary search trees. Given a BST T, how do we find the minimum and maximum elements present in the tree? Remember that if ‘y’ is in left subtree of node x, then key(y) is less than or equal to key(x). If ‘y’ is in right sub-tree of node x, then key(y) is greater than key (x). Since this property is satisfied at each node, the minimum element will be located at the leftmost node of the tree. The maximum element will be located at the rightmost node of the tree.

Given below is the pseudo-code for functions FindMinimumInBST and FindMaximumInBST. These functions take node ‘X’ which is the root of the BST.

FindMinimumInBST(X)
{
     while (left(X) != NULL)
          X = left(X);
     return X;
}
FindMaximumInBST(X)
{
     while (right(X) != NULL)
          X = right(X);
     return X;
}

What is the worst case time complexity of FindMinimumInBST and FindMaximumInBST? These functions traverse the BST from the root to the leaf. Hence their complexity is of the order of O(h) where ‘h’ is the height of the BST. Now the interesting question is that, given a binary search tree with N elements, what is the best possible height of the tree and what is the worst-case value of the height of the tree? For a complete binary tree with N elements, the height is O (logN) and this is the best-case value for the height of the tree. The worst-case value occurs when the tree grows linearly in one direction. The worst case value is O(N). I leave it to the reader to come up with examples for the order of tree construction such that we end up with trees of heights O(logN) and O(N).

BSTs support many dynamic set operations such as insert, delete and find. They can be used for implementing dictionaries and priority queues. Since we have already seen how to find the maximum and minimum, I leave it to the reader to come up with the code for ‘search’, ‘insert’ and ‘delete’. For insertion and deletion, you have to ensure that the BST property holds good after the operation. All basic operations take time, proportional to the height of the tree. Since the height of the tree in the best-case is O(logN), the best-case time for these operations is O(logN). Since the height of the tree in worst-case is O(N), the worst-case time for insert, search and delete operations is O(N). Hence in the worst case, the binary search tree is no better than a simple linked list for supporting these operations. There are various data structures like AVL trees, which are balanced trees. They guarantee a worst-case time of O(logN) for ‘insert’, ‘delete’ and ‘search’ operations by restructuring the tree so that the height remains O(logN). We shall discuss them in next month’s column.

We next look at another easy problem in the context of BSTs, that of finding the successor and predecessor of a given node. Given a node X with key value equal to key[X], the successor of X is the node Y such that key[Y] is the smallest key greater than key[X]. If node X is the maximum node in the binary search tree, then X has no successor. Note that finding the successor and predecessor requires no key comparisons. It requires only a tree traversal checking the tree structure.

If node ‘X’ has a non-empty right subtree, then the successor of X is the minimum node in X’s right subtree. If node X has an empty right subtree, X’s successor is the node Y for which X is found to be the predecessor—that is, X is the maximum in the left subtree of node Y. Given below is the pseudo-code for finding the successor of a given node X:

struct node FindSuccessor(struct node X)
{
     if (right[X] is != NULL)
     {
          return FindMinimumInBST(right[X]);
     }
     Y = parent[X];
     while ((Y != NULL)  && (X == right[Y]))
     {
          X = Y;
          Y = Parent[Y];
     }
     return Y ;
}

I leave it to the reader to come up with the code for FindPredecessor.

This month’s takeaway problem

This month’s takeaway problem comes from a reader, Rajeev Kumar. This is related to the inversions that form the basis for analysing the time complexity of sorting algorithms.

Inversion of a permutation is defined as follows:

  • Permutation: 6 2 3 1 4 7 8 9 5
  • Inversion: 3 1 1 1 4 0 0 0 0

Here, 1 has inversion 3, as there are three elements greater than 1 which are left to 1 in the given permutation, and these are 6, 2, and 3. Similarly, 2 has inversion 1 because there is only one element left to 2 which is greater than 2, that is, 6. The problem is to find inversion of a permutation in an efficient manner. It is easy to come with an algorithm with O(n^2) time complexity, but can we find inversion of a permutation in O(nlogn) time complexity even at the cost of some memory?
If you have any favourite programming puzzles that you would like to discuss on this forum, please send them to me. Feel free to send your solutions and feedback to me at sandyasm_AT_yahoo_DOT_com. Till we meet again next month, happy programming!

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