Given a binary tree, find the largest binary SEARCH tree in this binary tree.
When I mean largest, the tree with maximum number of nodes. And the binary search tree should be a sub-tree in the given binary tree.
Note: Not every binary tree is a binary search tree 😛
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any restrictions on running time of algo?
Worst case O(n2) might to be run isBST on each node from root to leaves..and the first instance of isBST returning true would be our tree right?
Try making it better. The better the complexity the better it is 🙂
Can we make it in O(n). where n is the number of the nodes in the given “binary tree”.
Hmm…In that case we may need to use extra space..and that standard solution of inorder traversal and finding max sorted subarray in it will help us…
Can you explain a bit clear… find the max sorted array would not help… consider the example
7
/ \
5 6
/ /
3 10
max sorted array in order traversal is 3 5 7 10, but this is not the answer.
hmm…yaa..totally missed that….thanks for correcting…
need more thought (atleast for me)
global list;
int fun(treenode node)
{
if(node == NULL)
{ return 0;
}
a = fun(node.left);
b= fun(node.right);
if( (a >=0) && (b >=0))
{
if(node.left == NULL)
{
if(node.right == NULL)
{
list.add(node, 1);
return 1;
}
else if(getfirst_node_inorder_traversal(node.right).value > = node.value)
{
list.add(b+1+a , node);
return b+a+1;//a will be zero as left is null
}
else
return -1;
}
else if(node.right == NULL)
{
if(getlast_node_inorder_traversal(node.left).value = node.value) && (getlast_node_inorder_traversal(node.left).value < = node.value) )
{
list.add(a+b+1,node);
return a+b+1;
}
else
return -1;
}
else
{
return -1;
}
}//end of fun
// Now from the list get the pair whose count is greater than all elements in the list. The corresponding node is the maximum binary search tree that we are looking.
The complexity for the above one will be O(n). where n is the number of elements present in the binary tree.
Just a rough glance says that this is not O(n) -> It would be O(nlogn)
And I havent verified your algo. Will do this in a spare time