Example 1 [Answer: O(n)]

public static void method(int n) { 
    int sum = 0;                    
    for (int i = 1; i <= n; i++) { 
        sum++;                    
        System.out.println(sum); 
    }
}    

Example 2 [Answer: O(n^2)]
    
public static void method(int n) {
    int sum = 0;                      
    for (int i = 1; i <= n; i++) {   
        for (int j = 1; j <= n; j++) { 
            sum++;                    
        }
    }
}

Example 3 [Answer: O(n)]

public static void method(int n) {
    int sum = 0;                       
    for (int i = 1; i <= n; i++) {    
        for (int j = 1; j <= 4; j++) {
            sum++;                   
        }
    }
}

Example 4 [Answer: O(n^2)]

public static void method(int n) {
    int sum = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            sum++;
        }
    }
}

Example 5 [Answer: O(n)]
Note: This answer requires developing and solving a recurrence relation
Recurrence Relation: 
      T(0) = 1
      T(n) = 1 + 2T(n/2)
Where n is the number of nodes in the tree with root current

// Pre-Condtion: Current is not null
// This algorithm only works for balanced binary trees
public static int countNodes(Node current) {
    if (current.left == null && current.right == null) {  
        return 1;                                        
    } else {
        int leftCount = countNodes(current.left);
        int rightCount = countNodes(current.right);
        return leftCount + rightCount + 1;
    }
}