Question

Find the median in a given BST with n nodes. The code should take O(H) time, where is the height of the BST. You can assume that the size of the tree is an odd number.

1. Describe your idea in short sentences below?
2. Provide the code that finds the median in the BST and rune to see if it works.

Class Node{

int data;

int size;//size of the subtree rooted at this node, including thi node.

Node left, right;

}

class BST{

Node root;

Int median(){

Int answer =0;

Return answer;//this contains the median in T

}

}//end class BST

Answer 1

IDEA:

1)first we will find inorder traversal of the BST.

2)then count the total number of nodes.

3)If nodes is even: then

median = (middle node+ very next node) /2

4)If nodes is odd : then

median = (very next node to middle node) node.

CODE:

Class Node{

int data;

int size;//size of the subtree rooted at this node, including thi node.

Node left, right;

}

class BST{

Node root;
static int counNodes(Node root)
{
Node current, pre;
  
  
int count = 0;
  
if (root == null)
return count;
  
current = root;
while (current != null)
{
if (current.left == null)
{
// Count node if its left is NULL
count++;
  
// Move to its right
current = current.right;
}
else
{
/* Find the inorder predecessor of current */
pre = current.left;
  
while (pre.right != null &&
pre.right != current)
pre = pre.right;
  
/* Make current as right child of its
inorder predecessor */
if(pre.right == null)
{
pre.right = current;
current = current.left;
}
else
{
pre.right = null;
  
// Increment count if the current
// node is to be visited
count++;
current = current.right;
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
  
return count;
}

Int median(node root){

Int answer =0;
if (root == null)
return 0;
  
int count = counNodes(root);
int currCount = 0;
Node current = root, pre = null, prev = null;
  
while (current != null)
{
if (current.left == null)
{
// count current node
currCount++;
  
// check if current node is the median
// Odd case
if (count % 2 != 0 && currCount == (count+1)/2) {
answer=prev.data;   
return answer;
}
// Even case
else if (count % 2 == 0 && currCount == (count/2)+1)
{ answer= (prev.data + current.data)/2;
return answer;
} // Update prev for even no. of nodes
prev = current;
  
//Move to the right
current = current.right;
}
else
{
/* Find the inorder predecessor of current */
pre = current.left;
while (pre.right != null && pre.right != current)
pre = pre.right;
  
/* Make current as right child of its inorder predecessor */
if (pre.right == null)
{
pre.right = current;
current = current.left;
}
  
/* Revert the changes made in if part to restore the original
tree i.e., fix the right child of predecssor */
else
{
pre.right = null;
  
prev = pre;
  
// Count current node
currCount++;
  
// Check if the current node is the median
if (count % 2 != 0 && currCount == (count+1)/2 )
{ answer= current.data;
return answer;
}
else if (count%2==0 && currCount == (count/2)+1)
{ answer= (prev.data+current.data)/2;
return answer;
}
// update prev node for the case of even
// no. of nodes
prev = current;
current = current.right;
  
} /* End of if condition pre->right == NULL */
} /* End of if condition current->left == NULL*/
} /* End of while */
return -1;

}

}//end class BST