info FindBst(Node root)
{
if (root == null)
{
return(new info(0, int.MinValue, int.MaxValue, 0, true));
}
if (root.left == null && root.right == null)
{
return(new info(1, root.data, root.data, 1, true));
}
info left = FindBst(root.left);
info right = FindBst(root.right);
info ret;
int size = 1 + left.size + right.size;
int min = left.min;
int max = right.max;
//Tree inculuding root is BST when below condition is true
if (left.isbst && right.isbst && left.max < root.data && right.min > root.data)
{
ret = new info(size, min, max, size, true);
return(ret);
}
//Either left or right subtree of root is BST hence return the subtree(left or right) which has largest size
ret = new info(size, min, max, Math.Max(left.actual_size, right.actual_size), false);
return(ret);
}
public largest_bst_in_given_binary_tree()
{
/*
* Input:
* 5
* / \
* 2 4
* / \
* 1 3
*
* Output: 3
* The following subtree is the
* maximum size BST subtree
* 2
* / \
* 1 3
*/
//Node root = new Node(5);
//root.left = new Node(2);
//root.left.left = new Node(1);
//root.left.right = new Node(3);
//root.right = new Node(4);
/*
* Given a Binary Tree, write a function that returns the size of the largest subtree which is also a Binary Search Tree (BST). If the complete Binary Tree is BST, then return the size of whole tree.
*
* Examples:
*
* Input:
* 5
* / \
* 2 4
* / \
* 1 3
*
* Output: 3
* The following subtree is the maximum size BST subtree
* 2
* / \
* 1 3
*
*
* Input:
* 50
* / \
* 30 60
* / \ / \
* 5 20 45 70
* / \
* 65 80
* Output: 5
* The following subtree is the maximum size BST subtree
* 60
* / \
* 45 70
* / \
* 65 80
*
*
*/
Node root = new Node(50);
root.left = new Node(30);
root.right = new Node(60);
root.left.left = new Node(2);
root.left.right = new Node(20);
root.right.left = new Node(45);
root.right.right = new Node(70);
root.right.right.left = new Node(65);
root.right.right.right = new Node(80);
info obj = FindBst(root);
}
