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CSE143 Key to Sample Final (Part 2) handout #34
1. Binary Trees, 10 points. One possible solution appears below.
public int purebredOdds() {
return purebredOdds(overallRoot);
}
private int purebredOdds(IntTreeNode root) {
if (root == null || root.data % 2 == 0)
return 0;
else
return 1 + purebredOdds(root.left) + purebredOdds(root.right);
}
2. Comparable class, 20 points. One possible solution appears below.
public class FoodData implements Comparable<FoodData> {
private String name;
private double fat;
private double carbs;
private double protein;
public FoodData(String name, double fat, double carbs,
double protein) {
if (fat < 0 || carbs < 0 || protein < 0)
throw new IllegalArgumentException();
this.name = name;
this.fat = fat;
this.carbs = carbs;
this.protein = protein;
}
public String getName() {
return name;
}
public double getCalories() {
return 9 * fat + 4 * (carbs + protein);
}
public double percentFat() {
if (getCalories() == 0)
return 0.0;
else
return 100.0 * (9 * fat / getCalories());
}
public String toString() {
return name + ": " + fat + "g fat, " + carbs +
"g carbohydrates, " + protein + "g protein";
}
public int compareTo(FoodData other) {
double difference = percentFat() - other.percentFat();
if (difference < 0)
return -1;
else if (difference > 0)
return 1;
else
return name.compareTo(other.name);
}
}
3. Binary Trees, 20 points. One possible solution appears below.
public void mirror() {
mirror(overallRoot);
}
private void mirror(IntTreeNode root) {
if (root != null) {
IntTreeNode temp = root.left;
root.left = root.right;
root.right = temp;
mirror(root.left);
mirror(root.right);
}
}
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