Recursion

Consider the following C functions.

int fun1(int n) {               int fun2(int n) {
  static int i = 0;                static int i = 0;
  if (n > 0) {                     if (n > 0) {
     ++i;                             i = i + fun1(n);
     fun1(n-1);                       fun2(n-1);
  }                                }
  return(i);                       return(i);
}                                }

The return value of fun2(5) is __________.

Consider the following function

double f(double x){
    if( abs(x*x - 3) < 0.01) return x;
    else return f(x/2 + 1.5/x);
}

Give a value q (to 2 decimals) such that f(q) will return q: _____.

Consider the C function given below.

int f(int j)
{
    static int i = 50;
    int k;
    if (i == j)
    {
        printf("something");
        k = f(i);
        return 0;
    }
    else return 0;
}

Which one of the following is TRUE?

Consider the pseudocode given below. The function DoSomething() takes as argument a pointer to the root of an arbitrary tree represented by the leftMostChild-rightSibling representation. Each node of the tree is of type treeNode.

typedef struct treeNode* treeptr;

struct treeNode
{
    treeptr leftMostChild, rightSibling;
};

int DoSomething (treeptr tree)
{
    int value=0;
    if (tree != NULL) {
        if (tree->leftMostChild == NULL)
            value = 1;
        else
            value = DoSomething(tree->leftMostChild);
        value = value + DoSomething(tree->rightSibling);
    }
    return(value);
}

When the pointer to the root of a tree is passed as the argument to DoSomething, the value returned by the function corresponds to the

The height of a tree is defined as the number of edges on the longest path in the tree. The function shown in the pseudocode below is invoked as height(root) to compute the height of a binary tree rooted at the tree pointer root.

int height (treeptr n)
{ if (n == NULL) return -1;
  if (n -> left == NULL)
     if (n -> right == NULL) return 0;
     else return [B1] ;                    // Box 1
  else { h1 = height (n -> left);
         if (n -> right == NULL) return (1+h1);
         else { h2 = height (n -> right);
                return [B2] ;              // Box 2
              }
       }
}

The appropriate expressions for the two boxes B1 and B2 are