Generated by Codex with GPT-5

Quick facts

What the problem is really asking

The task is to read a binary tree one layer at a time.

Instead of visiting nodes in a depth-first order such as root, left subtree, right subtree, the problem wants the tree grouped by depth:

  • level 0: the root
  • level 1: the root’s children
  • level 2: the grandchildren
  • and so on

So the real job is not just “traverse the tree.” It is “traverse the tree while keeping nodes from the same depth together.”

The key idea

The most natural way to process a tree level by level is breadth-first search.

A queue gives exactly the ordering the problem needs. When a node is removed from the queue, its children are appended to the back. That means all nodes already in the queue belong to the current level or the next one, and if the algorithm snapshots the queue length before starting a loop, it knows exactly how many nodes belong to the current level.

That one detail, capturing level_size = len(queue), is what turns a normal BFS into a grouped level-order traversal.

How to derive the optimal solution

Start from the output format. The answer is a list of lists, where each inner list contains all node values at one depth.

That immediately suggests two requirements:

  1. nodes must be visited in increasing depth order
  2. nodes from the same depth must be collected together before moving on

Depth-first search handles the second part if the algorithm records the current depth, but breadth-first search handles both parts almost automatically. The queue already visits shallower nodes before deeper ones. The only extra work is to process the queue in fixed-size batches so each batch becomes one output row.

That leads to the standard approach:

  1. return an empty list if the tree is empty
  2. initialize a queue with the root
  3. for each loop, record how many nodes are currently in the queue
  4. remove exactly that many nodes, collecting their values
  5. append each removed node’s children to the queue
  6. store the collected values as one finished level

This touches every node once, so the runtime is linear.

Best approaches

The queue-based BFS solution is the best default answer in an interview because it matches the problem statement directly. It runs in O(n) time, where n is the number of nodes, and uses up to O(w) queue space, where w is the maximum width of the tree.

A depth-first search solution is also valid and still O(n). In that version, the recursion or explicit stack keeps track of the current depth, and the algorithm appends each value into result[depth]. That is also optimal in asymptotic terms, but BFS is usually easier to explain for this exact question because “level order” is literally what BFS does.

Python solution

from collections import deque
from typing import Deque, List, Optional


# LeetCode provides the TreeNode class.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right


class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if root is None:
            return []

        nodes_to_visit: Deque[TreeNode] = deque([root])
        traversal_by_level: List[List[int]] = []

        while nodes_to_visit:
            traversal_by_level.append(self._collect_next_level(nodes_to_visit))

        return traversal_by_level

    def _collect_next_level(self, nodes_to_visit: Deque[TreeNode]) -> List[int]:
        level_size = len(nodes_to_visit)
        current_level_values: List[int] = []

        # Process exactly one level before any newly discovered children
        # are handled in the next outer-loop iteration.
        for _ in range(level_size):
            current_node = nodes_to_visit.popleft()
            current_level_values.append(current_node.val)

            if current_node.left is not None:
                nodes_to_visit.append(current_node.left)
            if current_node.right is not None:
                nodes_to_visit.append(current_node.right)

        return current_level_values

Why this works

At the start of each outer loop, the queue contains exactly the nodes that still need processing, ordered from shallowest to deepest. Because children are appended only after their parent is removed, every node from the current level appears before every node from the next level.

The algorithm records the queue length before processing that round. That number is exactly the number of nodes in the current level. By removing only that many nodes, it finishes one level cleanly and does not mix in the next one.

Every node enters the queue once and leaves the queue once, so the total work is O(n). The grouping is correct because the queue preserves breadth-first order and the fixed-size batch preserves level boundaries.

Interview follow-ups

How would you solve this with depth-first search instead?

The depth-first version keeps the current depth as an argument in the recursive call. When visiting a node at depth d, the algorithm first checks whether result already has a list for that depth. If not, it appends a new empty list. Then it adds the node’s value into result[d] and recurses into the left and right children with depth d + 1.

This works because every node knows its own depth from the recursion path, so the algorithm can still group values by level even though the traversal order is not breadth-first. The runtime stays O(n), but the extra space shifts from a queue to the recursion stack. In a very unbalanced tree, that stack can grow to O(h), where h is the tree height. It is a good alternative, but BFS is still the more direct fit for the original question.

What changes if the output should be bottom-up level order?

Almost nothing changes in the traversal itself. The algorithm can still perform the same BFS and collect levels from top to bottom. After that, it can reverse the final list once before returning it.

That approach works because the grouping logic is unchanged; only the final presentation order changes. The full runtime is still O(n), plus O(L) for reversing L levels, which is already bounded by O(n). Another option is to prepend each level as it is produced, but that is usually less clean in Python because front insertion into a normal list is more expensive than appending and reversing once at the end.

How would you return the zigzag level order traversal?

The same level-by-level BFS structure still works. The only difference is how each finished level is written into the result. One simple method is to keep a boolean flag such as left_to_right. For normal levels, append values as usual. For reversed levels, either reverse the collected level before storing it or build it with a deque and insert values on the opposite side.

This works because the tree is still processed one level at a time; only the order within each level changes. If the code reverses a list after collecting it, the solution remains easy to read and stays O(n) overall. Using a deque for each level avoids a reversal step, but it adds a little more implementation detail. In interviews, the append-then-reverse version is often the clearest answer unless the interviewer asks to optimize constant factors.