Pattern this teaches:Tree BFS and Level Order
Binary Tree Level Order Traversal
A medium Binary Trees interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Binary Trees
- Sheets
- 3
- Core for
- 4 roles
The problem
Read a binary tree level by level, from the root downwards, and within each level from left to right. Return one list of values per level.
Example 1
- Input
- [5,3,7,1,4,6,8]
- Output
- [[5],[3,7],[1,4,6,8]]
- Why
- Level 0: 5, Level 1: 3,7, Level 2: 1,4,6,8.
Example 2
- Input
- [1]
- Output
- [[1]]
- Why
- A single node forms one level with just that node.
Example 3
- Input
- []
- Output
- []
- Why
- An empty tree has no levels.
Constraints
- The number of nodes is in the range [0, 2000].
- -1000 <= Node.val <= 1000
How to think about it
Updated 2026-09-29Level order is an exact FIFO frontier. To chunk nodes into distinct sub-arrays per depth, freeze the size of the queue before pulling any elements for that round. Every node popped within that count belongs to the current level, while every child pushed belongs strictly to the next.
Approaches, worst first
DFS with depth indexing
time O(n) · space O(h)
Pass depth down recursive calls. If `result.length === depth`, insert a new empty list, then append the node's value to `result[depth]`. Recursion handles depth grouping without maintaining an active frontier.
BFS with batch level poppingWrite this one
time O(n) · space O(n)
Initialize a queue with the root. While the queue is not empty, capture its length `k`, pop exactly `k` nodes into a row list, and enqueue their non-null children. Push each completed row into the answer list.
Where people lose marks · 3
- Re-evaluating `queue.length` dynamically inside the level loop causes newly enqueued children to be processed in the current level instead of the next.
- Using array shift in languages where `shift()` runs in linear time can degrade total runtime to quadratic if a true queue or index offset is not used.
- Failing to check for an empty root returns `[[]]` instead of `[]`.
The theory behind it
Binary Trees — the ground this problem stands on. All Binary Trees problems
What Binary Trees is
A binary tree is a branching data structure that starts at a single top node called the root, like an upside-down family tree. Every node holds a piece of data and can branch out to at most two children below it, known as the left child and the right child. Because there is no ordering rule about which values go left or right, finding a specific item can require checking every single node in the entire tree.
When to reach for it
Reach for binary trees when problems present hierarchical data with left and right child pointers. Questions asking for tree height, maximum depth, path sums from root to leaf, diameter, lowest common ancestor, or checking whether two trees are mirror reflections of each other all signal binary tree traversals. Any problem asking to inspect or reconstruct a tree layer by layer or path by path belongs here.
How the pattern works
Think recursively by focusing on what a single node must do. If the current node is null, return the base answer immediately. Otherwise, ask the left child for its result, ask the right child for its result, and combine both answers with the current node value before returning up to the parent. For horizontal scans, use a queue to read nodes layer by layer, measuring the queue length at the start of each layer to group nodes by depth.
More Binary Trees problems
Target Roles
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