Binary Tree Level Order Traversal II

Medium Trees & BST Traversals Original on LeetCode

Binary Tree Level Order Traversal II is a medium trees & bst problem solved with the traversals pattern. The best approach, bfs, adding each level at the front, runs in O(n) time and O(w) space. Below are 2 approaches in Java, from bfs, then reverse the list of levels up.

Problem

Return the level order traversal from the bottom level up to the root, left to right within each level.

Examples

Example 1

Input: root = [3, 9, 20, null, null, 15, 7]
Output: [[15, 7], [9, 20], [3]]

Example 2

Input: root = [1]
Output: [[1]]

Constraints

The tree has 0 to 2000 nodes.

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
BFS, then reverse the list of levels	`O(n)`	`O(w)`
BFS, adding each level at the front	`O(n)`	`O(w)`

1BFS, then reverse the list of levels

TimeO(n)

SpaceO(w)

Run normal level order and reverse the outer list at the end.

Standard BFS; Collections.reverse(out).

Java

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> out = new ArrayList<>();
        if (root == null) return out;
        Queue<TreeNode> q = new ArrayDeque<>();
        q.add(root);
        while (!q.isEmpty()) {
            List<Integer> level = new ArrayList<>();
            for (int i = q.size(); i > 0; i--) {
                TreeNode n = q.poll();
                level.add(n.val);
                if (n.left != null) q.add(n.left);
                if (n.right != null) q.add(n.right);
            }
            out.add(level);
        }
        Collections.reverse(out);
        return out;
    }
}

2BFS, adding each level at the front

TimeO(n)

SpaceO(w)

Insert each level at index 0 of a LinkedList, so the deepest level ends up first with O(1) insertion.

out = new LinkedList; out.addFirst(level) after each level.

Java

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        LinkedList<List<Integer>> out = new LinkedList<>();
        if (root == null) return out;
        Queue<TreeNode> q = new ArrayDeque<>();
        q.add(root);
        while (!q.isEmpty()) {
            List<Integer> level = new ArrayList<>();
            for (int i = q.size(); i > 0; i--) {
                TreeNode n = q.poll();
                level.add(n.val);
                if (n.left != null) q.add(n.left);
                if (n.right != null) q.add(n.right);
            }
            out.addFirst(level);
        }
        return out;
    }
}

Edge cases to test

Empty tree

Hints

Hint 1

Same BFS; only the order you add levels to the answer changes.

FAQ

What is the best time complexity for Binary Tree Level Order Traversal II?

BFS, adding each level at the front runs in O(n) time and O(w) extra space.

Which pattern does Binary Tree Level Order Traversal II use?

It is a trees & bst problem that uses the traversals pattern. Other problems with the same pattern: Binary Tree Level Order Traversal, Binary Tree Zigzag Level Order Traversal.

Is there a brute force solution for Binary Tree Level Order Traversal II?

Yes. BFS, then reverse the list of levels takes O(n) time and O(w) space. Run normal level order and reverse the outer list at the end.

Which edge cases should I test for Binary Tree Level Order Traversal II?

Empty tree.