Question

in C++, given a binary search tree of ints, in which each node contains a size parameter (also an int), explain, in English, not code, how you would find the median element of the tree in theta(log N) time.

Answer 1

`Hey,

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We can find kth element in BST where k=n/2 for median

Here's just an outline of the idea:

In a BST, the left subtree of node T contains only elements smaller than the value stored in T. If k is smaller than the number of elements in the left subtree, the kth smallest element must belong to the left subtree. Otherwise, if k is larger, then the kth smallest element is in the right subtree.

We can augment the BST to have each node in it store the number of elements in its left subtree (assume that the left subtree of a given node includes that node). With this piece of information, it is simple to traverse the tree by repeatedly asking for the number of elements in the left subtree, to decide whether to do recurse into the left or right subtree.

Now, suppose we are at node T:

1. If k == num_elements(left subtree of T), then the answer we're looking for is the value in node T.
2. If k > num_elements(left subtree of T), then obviously we can ignore the left subtree, because those elements will also be smaller than the kth smallest. So, we reduce the problem to finding the k - num_elements(left subtree of T) smallest element of the right subtree.
3. If k < num_elements(left subtree of T), then the kth smallest is somewhere in the left subtree, so we reduce the problem to finding the kth smallest element in the left subtree.

The abovee algorithm requires O(log(n))

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