I have written some code for finding a level in a binary tree, having a maximum number of elements. I have a few questions:
- Is it a good design? I have used 2 queues but the total sum of elements both queues store will be less than n. So I think it should be OK.
- Can there be a better design?
public class MaxSumLevel {
public static int findLevel(BinaryTreeNode root) {
Queue mainQ = new Queue();
Queue tempQ = new Queue();
int maxlevel = 0;
int maxVal = 0;
int tempSum = 0;
int tempLevel = 0;
if (root != null) {
mainQ.enqueue(root);
maxlevel = 1;
tempLevel = 1;
maxVal = root.getData();
}
while ( !mainQ.isEmpty()) {
BinaryTreeNode head = (BinaryTreeNode) mainQ.dequeue();
BinaryTreeNode left = head.getLeft();
BinaryTreeNode right = head.getRight();
if (left != null) {
tempQ.enqueue(left);
tempSum = tempSum + left.getData();
}
if (right != null) {
tempQ.enqueue(right);
tempSum = tempSum + right.getData();
}
if (mainQ.isEmpty()) {
mainQ = tempQ;
tempQ = new Queue();
tempLevel ++;
if (tempSum > maxVal) {
maxVal = tempSum;
maxlevel = tempLevel;
tempSum = 0;
}
}
}
return maxlevel;
}
}
int
)? Does it correspond to the depth of a node? If so, there’s a much simpler solution. \$\endgroup\$O(1)
function call. \$\endgroup\$