Binary Tree Level Order Traversal

Problem

Given the root of a binary tree, return its level-order traversal: a list of lists, one list of values per level (left to right).

Constraints

• 0 ≤ nodes ≤ 2000

Approach — Tree Traversal

This is a Tree Traversal problem. The idea: traverse the tree — DFS recursively, or BFS level by level — and combine the results from each subtree. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(n) time.

Solution code

Python

class Solution:
    def levelOrder(self, root):
        res = []
        level = [root] if root else []
        while level:
            res.append([n.val for n in level])
            level = [c for n in level for c in (n.left, n.right) if c]
        return res

Java

import java.util.*;

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> out = new ArrayList<>();
        if (root == null) return out;
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int sz = q.size();
            List<Integer> row = new ArrayList<>(sz);
            for (int i = 0; i < sz; i++) {
                TreeNode n = q.poll();
                row.add(n.val);
                if (n.left  != null) q.offer(n.left);
                if (n.right != null) q.offer(n.right);
            }
            out.add(row);
        }
        return out;
    }
}

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.