Starter Code Here is part of the Tree class def. You need to find out what's missing and fill it in." class Tree { private: struct Node { / / Inner class std: :string data; Node * sibling, *_child; static bool is_equal (const Node *pl, const Node *p2) ; Node (std: : string s = "") : // TODO Node (const Node &that) ; / / TODO const Node Goperator=(const Nodes that) ; // Deep clone ~Node () ; std: : string get_data () const { return data; } void set_data (std: :string s) { _data = s; } Node *insert_sibling (Node *p) ; Node *insert_child (Node *p) ; std: :string to_string () const; bool operator==(const Node &that) const; bool operator!=(const Node &that) const; Node * root; public: Tree () ; -Tree () ; Tree (const Trees that) { *this = that; } Trees operator=(const Trees that) ; // Deep clone std: :string to_string () const; void make_special_config 1 (const std: :vector& names) ;Your tenth miniquest - Tree copy Implement the copy constructor and the assignment operator, though not necessarily in that order. Tree (const Tree & that) ; Tree Soperator=(const Trees that) ; // Deep clone The assignment operator should clone the RHS tree (perform a deep copy). The copy constructor should simply invoke the assignment operator. Again, make sure to check for the special case when someone accidentally assigns a tree to itself. What would happen then if you were careless?Your eleventh miniquest - Tree comparisons Implement: bool operator== (const Tree &that) ; bool operator!=(const Tree &that) const; The == operator returns true if the two trees are structurally identical. Actual node pointers may be different. By structurally identical, I mean that they have the equal root nodes, where node equality is defined as in the earlier miniquest (I think 7). Your twelfth miniquest - Tree to string Implement: to_string () const; This is a very easy miniquest if you've managed to ace the Node : : to_string () miniquest. Otherwise... not so much. It is also an optional miniquest (return " to skip). Simply return the stringified version of the root wrapped within comment lines as shown below: # Tree rooted at [X] . # The following lines are of the form: # node: childl child2... [ . . .] + # End of Tree There is exactly one space before each word, except the word "node" on the last line where there are 3 spaces (gray rectangle). As before, the red X in square brackets must be replaced by the name of the root node (which should be ROOT if everything is set up right). . stands for a newline.Your thirteenth miniquest - Special tree This is a fun miniquest. All you have to do is to construct a Tree that looks like Fig 3. That's all. Implement the method: void make_special_config_1 (const vector & names) ; bool operator== (const Tree &that) const { // TODO bool operator!= (const Tree that) const { / / TODO friend std: : ostreams operator