Homework Answers
A)
public static <AnyType> int
evenElements(BinaryNode<AnyType> t) {
if(t==null)
return 0;
int value = 0;
if(t.getElement() %2 ==0 )
value = 1;
return value + evenElements(t.getLeft()) +
evenElements(t.getRight());
}
Big Oh running time is O(n) where n is the number of elements in
the Tree. Since we need to visist all the element atleast once to
determine the even/odd. Hence time complexity is O(n)
B)
public static <AnyType> int
sumElements(BinaryNode<AnyType> t) {
if(t==null)
return 0;
return t.getElement() + sumElements(t.getLeft()) +
sumElements(t.getRight());
}
Big Oh running time is O(n) where n is the number of elements in
the Tree. Since we need to visit all the element atleast once to
determine the sum. Hence time complexity is O(n)
Thanks, PLEASE COMMENT if there is any concern.
i am confuse this answer and i am new this website please can you help me answer 10E or 18.10 b
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