Consider a binary tree of integers dynamically implemented; write a function update that changes node values such that each node is equal to the sum of its children. It is obvious that leaves remain unchanged. 

Example:

void update(Btree *B)
{
	if(*B==NULL || (*B)->left==NULL && (*B)->right==NULL) return ;
	update(&((*B)->left));
	update(&((*B)->right));
	if((*B)->left && (*B)->right)
		(*B)->data = (*B)->left->data  + (*B)->right->data;
	else
		if((*B)->left)
			(*B)->data = (*B)->left->data ;
		else
			if((*B)->right)
				(*B)->data = (*B)->right->data;
}