I am to write a function that receives the root of a binary tree t and a number k and determine if the node where k is stored is the left child of its parent. if k is not in the tree, or does not have a parent, or is a right child the function should return false. Starting code: def is_left_child(t,k):    parent, x = t.find(k, return_parent = True)    return False

Suppose a node of a linked list is defined as follows in your program:    typedef struct{  int data;  struct Node* link;  } Node;    Write the definition of a function, printOdd(), which displays all the odd numbers of a linked list. What will be the time complexity of printOdd()? Express the time complexity in terms of Big-O notation.

Implement a __setitem__ function that also supports negative indices. For example: W = L2(Node(10, Node(20, Node(30))))print W[ 10, 20, 30 ] W[1]=25print W[ 10, 25, 30 ] W[-1]=35print W[ 10, 25, 35 ] Complete the code: def L2(*args,**kwargs):         class L2_class(L):                 def __getitem__(self, idx):                         <... YOUR CODE HERE ...>                 def __setitem__(self, idx, value):            <... YOUR CODE HERE ...>                 return L2_class(*args,**kwargs) W = L2(Node(10, Node(20, Node(30))))print(W)W[1]=25print(W)W[-1]=35print(W)

What statement is the most appropriate for the following f1 function?   template <class Item> int f1(binary_tree_node<Item> node) {      int answer = 0;      if (node->left( )!= NULL) ++answer;      if (node->right( )!= NULL) ++answer;       eturn answer; } Group of answer choices The f1 function computes the number of left children that a node has. The f1 function computes the number of ancestors that a node has. The f1 function computes the number of leaves that a node has. The f1 function computes the number of children that a node has. The f1 function computes the number of right children that a node has.