binary-trees-answers.md

CSE 143

Binary Trees Answers

Final Study Session

1. evenBranches. One possible answer is shown below.

   public int evenBranches() {
   	return evenBranches(overallRoot);
   }
   
   private int evenBranches(IntTreeNode root) {
   	if (root == null) {
   		return 0;
   	} else if (root.left == null && root.right == null) {
   		return 0;
   	} else if (root.data % 2 == 0) {
   		return 1 + evenBranches(root.left) + evenBranches(root.right);
   	} else {
   		return evenBranches(root.left) + evenBranches(root.right);
   	}
   }

2. countMultiples. One possible answer is shown below.

   public int countMultiples(int value) {
   	if (value == 0) {
   		throw new IllegalArgumentException();
   	}
   	return countMultiples(overallRoot, value);
   }
   
   private int countMultiples(IntTreeNode root, int value) {
   	if (root == null) {
   			return 0;
   	} else {
   		int sum = countMultiples(root.left, value) + countMultiples(root.right, value);
   		if (root.data % value == 0) {
   			return 1 + sum;
   		} else {
   			return sum;
   		}
   	}
   }

3. pairsOfTwins. One possible answer is shown below.

   public int pairsOfTwins() {
   	return pairsOfTwins(overallRoot);
   }
   
   private int pairsOfTwins(IntTreeNode node) {
   	if (node == null)
   		return 0;
   	else if (node.left != null && node.right != null &&
   			node.left.data == node.right.data)
   		return 1 + pairsOfTwins(node.left) + pairsOfTwins(node.right);
   	else
   		return pairsOfTwins(node.left) + pairsOfTwins(node.right);
   }