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Given a binary tree, return a list of each level.

I'm looking for a general review and a mention on best-practices, optimization, and verification of my complexities.

Time complexity: \$O(n)\$

Space complexity: \$O(2^{height})\$

public class PrintEachLevel<T> {

    private TreeNode<T> root;

    public PrintEachLevel(List<T> items) {
        create(items);
    };

    private static class TreeNode<T> {
        TreeNode<T> left;
        T item;
        TreeNode<T> right;
        TreeNode(TreeNode<T> left, T item, TreeNode<T> right) {
            this.left = left;
            this.item = item;
            this.right = right;
        }
    }

    private void create (List<T> items) {
        root = new TreeNode<T>(null, items.get(0), null);

        final Queue<TreeNode<T>> queue = new LinkedList<>();
        queue.add(root);

        final int half = items.size() / 2;

        for (int i = 0; i < half; i++) {
            if (items.get(i) != null) {
                final TreeNode<T> current = queue.poll();
                final int left = 2 * i + 1;
                final int right = 2 * i + 2;

                if (items.get(left) != null) {
                    current.left = new TreeNode<T>(null, items.get(left), null);
                    queue.add(current.left);
                }
                if (right < items.size() && items.get(right) != null) {
                    current.right = new TreeNode<T>(null, items.get(right), null);
                    queue.add(current.right);
                }
            }
        }
    }



    public List<List<T>> getLevels() {
        if (root == null) {
            throw new NullPointerException("The root cannot be null.");
        }

        List<List<T>> levels = new ArrayList<>();
        List<T> level = new ArrayList<>();

        Queue<TreeNode<T>> queue = new LinkedList<>();
        Queue<TreeNode<T>> queueNext = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            TreeNode<T> node = queue.poll();
            level.add(node.item); 

            if (node.left != null) queueNext.add(node.left);
            if (node.right != null) queueNext.add(node.right);

            /**
             * http://stackoverflow.com/questions/24790507/to-swap-or-create-new-references
             */
            if (queue.isEmpty()) {
                queue = queueNext;
                queueNext = new LinkedList<>();
                levels.add(level);
                level  = new ArrayList<>();
            }
        }

        return levels;
    }
}


public class PrintEachLevelTest {

    /**
     * Simple tree with just 1 node
     */
    @Test
    public void test1() {
        PrintEachLevel<Integer> bfsTraversal1 = new PrintEachLevel<>(Arrays.asList(1));
        List<List<Integer>> listOflist1 = new ArrayList<>();
        listOflist1.add(Arrays.asList(1));
        assertEquals(listOflist1, bfsTraversal1.getLevels());
    }

    /** 
     *        10
     *      /    \
     *     5       18
     *    / \     / \
     *   4   6   17   19
     */
    @Test
    public void test2() {
        PrintEachLevel<Integer> bfsTraversal2 = new PrintEachLevel<>(Arrays.asList(10, 5, 18, 4, 6, 17, 19));
        List<List<Integer>> listOflist2 = new ArrayList<>();
        listOflist2.add(Arrays.asList(10));
        listOflist2.add(Arrays.asList(5, 18));
        listOflist2.add(Arrays.asList(4, 6, 17, 19));
        assertEquals(listOflist2, bfsTraversal2.getLevels());
    }

    /**
     *               1
     *           /      \
     *      null          2
     *     /   \         /   \
     *   null  null    null    3
     */
    @Test
    public void test3() {
        PrintEachLevel<Integer> bfsTraversal3 = new PrintEachLevel<>(Arrays.asList(1, null, 2, null, null, null, 3));
        List<List<Integer>> listOflist3 = new ArrayList<>();
        listOflist3.add(Arrays.asList(1));
        listOflist3.add(Arrays.asList(2));
        listOflist3.add(Arrays.asList(3));
        assertEquals(listOflist3, bfsTraversal3.getLevels());
    }

    /**
     *                 4
     *             /     \ 
     *          2          null
     *       /   \        /    \
     *      1    null  null    null
     */
    @Test
    public void test4() {
        PrintEachLevel<Integer> bfsTraversal4 = new PrintEachLevel<>(Arrays.asList(4, 2, null, 1, null));
        List<List<Integer>> listOflist4 = new ArrayList<>();
        listOflist4.add(Arrays.asList(4));
        listOflist4.add(Arrays.asList(2));
        listOflist4.add(Arrays.asList(1));
        assertEquals(listOflist4, bfsTraversal4.getLevels());
    }

    /**
     *             1
     *     2            3      
     *  4     null   null    7 
     * 
     */
    @Test
    public void test5() {
        PrintEachLevel<Integer> bfsTraversal5 = new PrintEachLevel<>(Arrays.asList(1, 2, 3, 4, null, null, 7));
        List<List<Integer>> listOflist5 = new ArrayList<>();
        listOflist5.add(Arrays.asList(1));
        listOflist5.add(Arrays.asList(2, 3));
        listOflist5.add(Arrays.asList(4, 7));
        assertEquals(listOflist5, bfsTraversal5.getLevels());
    }
}
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I think that this code is overly complex for the task on hand. Since you pass in a list that is already tree-like, there is no real reason to convert it into a real tree data structure and then back again. By eliminating the intermediate step, you can reduce overhead and increase speed tremendously.

Level 0 (the top node) starts at element 0
Level 1 starts at element 1
Level 2 starts at element 3
Level 3 starts at element 7
Level 4 starts at element 15
...
Level n starts at element 2^n - 1 and ends at element 2 * (2^n - 1)

Therefore to return any given level: iterate between the indices indicated which can be quickly calculated by some bit-shifting:

start: (2 << n) - 1 end: start << 1

Note: make sure to deal with the nulls while iterating in a desired way.


Since you want all the lists use the following code:

// create bounds
int size = items.size();
if (size == 0) return;

List<Integer> bounds = ArrayList<Integer>();
bounds.add(0);
bounds.add(1);
int last = 1;
while (size >= bounds.get(last)) {
    bounds.add(2 * bounds.get(last));
    last++;
}
if (bounds.get(last) != size) {
    bounds.remove(last);
    bounds.add(size);
}

// iterate through
List<List<Integer>> levels = new ArrayList<List<Integer>>();
for (int i = 0; i < bounds.size() - 1; i++) {
    int lower_bound = bounds[i];
    int upper_bound = bounds[i+1];
    List<Integer> level = ArrayList<Integer>();
    for (int j = lower_bound; j < upper_bound; j++) {
        level.add(items.get(j));
    levels.add(level);

return levels;
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  • \$\begingroup\$ this is an interview question - you have to deal in trees \$\endgroup\$ – JavaDeveloper Jul 17 '14 at 21:47
  • \$\begingroup\$ Well, this code would work as soon as you convert your tree into a flattened list. That is what you were passing your function anyways. If you were passing in a true tree structure (where you only pass the parent node), then that would be a completely different question. \$\endgroup\$ – mleyfman Jul 17 '14 at 21:49
  • \$\begingroup\$ interviewers would probably say you have a larger space complexity. these things dont matter much in real life but interviews - are based on all tiny optimizations \$\endgroup\$ – JavaDeveloper Jul 17 '14 at 21:56
  • \$\begingroup\$ If you are talking about creating the bounds array, those could be made programmatically as indicated above (which would make this algorithm utilize only as much space as the tree), but I choose to do it this way to make the loop look nice. The bounds array also only uses log(n) space, which is tiny when compared to the tree. \$\endgroup\$ – mleyfman Jul 17 '14 at 22:04
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A minor bug: You get an IndexOutOfBoundsException for an empty list in PrintEachLevel.create(List<T> items). You don't have a comment stating you need to input a list containing at least something. Consider returning IllegalArgumentException and adding a comment.

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