In object-oriented programming, each function is part of an object. The function can refer to the object in which it resides implicitly, so that object is not an explicit parameter. It is an implicit parameter.
A destructor recovers resources owned by an object. Typically,
it deletes memory that the object owns, but that is physically
outside the object.
The destructor for class River is called ~River.
It takes time proportional to nlog2(n).
The class, and private members of classes.
Even good programmers make mistakes, and the compiler should
catch those mistakes wherever possible.
Programming teams often have some less experienced programmers
who are likely to violate abstractions if permitted to do so.
A compiler that disallows such violations gives the better
programmers confidence that the less experienced programmers
are not violating abstractions.
81 / \ 20 100 / \ 16 65
81 65 / \ / \ 20 100 20 81 / \ / \ \ 16 65 16 25 100 / 25 before rotations after rotationsThe answer is the tree on the right. It is obtained from the tree on the left by doing a double rotation at the root.
25 / \ 10 30 / \ 26 50
Yes.
Yes.
25 / \ 10 30 / \ 26 50 / 40
No.
i Rep[i] i Rep[i] i Rep[i] i Rep[i] 1 1 1 _5__ 1 _5__ 1 _5__ 2 2 2 _2__ 2 _2__ 2 _4__ c(1,5) c(5,2) c(5,4) 3 3 --------> 3 _3__ --------> 3 _3__ --------> 3 _3__ 4 4 4 _4__ 4 _4__ 4 _4__ 5 5 5 _5__ 5 _2__ 5 _2__
struct TreeNode { int data; TreeNode* left; TreeNode* right; }; typedef TreeNode* Tree;
void PrintEvens(Tree T) { if(T != NULL) { PrintEvens(T->right); if(T->data % 2 == 0) cout << T->data << endl; PrintEvens(T->left); } }
Use the types TreeNode and Tree of the previous problem. Suppose that function node(x,lft,rgt) is available, where node(x,lft,rgt) returns a pointer to a newly allocated node holding data x, left subtree lft and right subtree rgt. (Do not write this function, just use it.)
Write a function removeMin(t) that returns the tree that would result from removing the smallest value from tree t, but that does not alter tree t. If t is empty, then removeMin(t) should return an empty tree. For example, if t is the tree
81 / \ 20 100 \ 65 / \ 50 70then removeMin(t) should return the following tree, without altering tree t.
81 / \ 65 100 / \ 50 70Do not rebalance the tree. The new tree that you construct can share subtrees with t, as long as it does not change the subtrees.
Tree removeMin(Tree t) { if(t == NULL) return NULL; else if(t->left == NULL) return t->right; else return node(t->data, removeMin(t->left), t->right); }