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A v

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A first of all we traverse the left node and print theis ..
leaf node funtion print all leaf node
right_node function print all right node .......

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v excellent ans

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R

void anticlock_edge(struct node *ptr)
{
	if(root==NULL)
	{
		printf("Tree is empty");
		return;
	}
	if(ptr!=NULL)
	{
		anticlock_edge(ptr->rchild);
		anticlock_edge(ptr->lchild);
		if(ptr->lchild==NULL

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S if the hight of the tree is' h'
then the no. of boundry nodes is=2*h-1