What is the pricing and subscription model for Hi👋SpeedAlgo?

The entire core curriculum of 13 patterns and 225 practice problems is 100% free to access with no credit card required. New accounts get 40 free coins once, the daily claim adds +20 coins once a day, and inviting a friend earns you +25 coins when they complete their first practice step. These are the only ways to earn coins. Each AI coach question costs 1-5 coins, priced by the server according to its size. AI coaching is free with your own Google Gemini, OpenAI or OpenRouter API key (BYOK). Coins also unlock extras for 24 hours: a Recall hint level (1 coin), a Solve solution reveal (2), a LeetCode problem card's solutions (1) and a pattern's extra canonical problems (1). Nothing on the site is sold for money: coins can only be earned, never bought.

What is the refund and cancellation policy?

Hi👋SpeedAlgo sells nothing for money: there are no subscriptions, renewals or charges, so there is nothing to cancel or refund. Nothing on the site is sold for money: coins can only be earned, never bought.

Who creates and maintains the curriculum?

Hi👋SpeedAlgo is developed and maintained by Hi👋WebEnterprise (https://hiwebenterprise.com), an engineering organization focused on developer education, deliberate practice methodologies, and cognitive ergonomics for technical interviews.

How does Hi👋SpeedAlgo prepare candidates for senior and staff SWE interviews?

Rather than rote LeetCode grinding, Hi👋SpeedAlgo uses deliberate practice for each pattern: learn the loop invariant, Big-O proof and edge cases from a side-by-side template and a step-by-step debugger, rebuild the template from memory, then solve a new problem on your own against a graded test suite.

1. Binary Tree Level Order Traversal • Invariant Proof & Speed-Run

Theory Context•Breadth-First Search (BFS)

MediumLC 102

Binary Tree Level Order Traversal (LeetCode 102)

Target Frequency:AmazonMetaGoogle

You have the root of a binary tree. Walk it breadth-first, visiting each depth in left-to-right order, and produce one list of values per depth — so the result is a list of lists, ordered from the root's level down to the leaves' level.

Worked Examples

Example 1

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

Output:[[3],[9,20],[15,7]]

Explanation: Depth 0 holds 3, depth 1 holds 9 and 20, depth 2 holds 15 and 7.

Example 2

Input:root = [1]

Output:[[1]]

Explanation: A single node forms one level.

Example 3

Input:root = []

Output:[]

Explanation: An empty tree has no levels.

⚖️Formal Constraints & Bounds

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

Deep-Dive & Conceptual Insights

Core Insight

Taking a snapshot of len(queue) separates nodes of the current level from their newly enqueued children.

Real-World Scenario & Production Applications

Peer-to-peer gossip network broadcasting and organizational hierarchy reporting charts.

Step-by-Step Execution Trace Table

Step-by-Step Wavefront Progression

Step 1 (Root Seed, Ring 0, d=0): queue = [3]. Seed queue with root node 3. Initial output = [].
Step 2 (Snapshot Ring 0, level_size=1, d=0): level_size = 1. Freeze Ring 0 boundary before child enqueue.
Step 3 (Dequeue 3 & Enqueue Children, Ring 0, d=0): Pop 3 $\rightarrow$ current_level = [3]. Enqueue left child 9 and right child 20 $\rightarrow$ queue = [9, 20]. Commit Ring 0 $\rightarrow$ output = [[3]] 📸.
Step 4 (Snapshot Ring 1, level_size=2, d=1): level_size = 2. Freeze Ring 1 boundary. queue = [9, 20].
Step 5 (Dequeue 9, Leaf, Ring 1, d=1): Pop 9 $\rightarrow$ current_level = [9]. Node 9 has no children. Remaining in Ring 1: 1 node.
Step 6 (Dequeue 20 & Enqueue Children, Ring 1, d=1): Pop 20 $\rightarrow$ current_level = [9, 20]. Enqueue left child 15 and right child 7 $\rightarrow$ queue = [15, 7]. Commit Ring 1 $\rightarrow$ output = [[3], [9, 20]] 📸.
Step 7 (Snapshot Ring 2, level_size=2, d=2): level_size = 2. Freeze Ring 2 boundary. queue = [15, 7].
Step 8 (Dequeue 15, Leaf, Ring 2, d=2): Pop 15 $\rightarrow$ current_level = [15]. Node 15 has no children. Remaining in Ring 2: 1 node.
Step 9 (Dequeue 7, Leaf, Ring 2, d=2): Pop 7 $\rightarrow$ current_level = [15, 7]. Node 7 has no children. Commit Ring 2 $\rightarrow$ output = [[3], [9, 20], [15, 7]] 📸.
Step 10 (Queue Empty, Complete, d=2): queue = []. All concentric level rings drained. Search Complete: return [[3], [9, 20], [15, 7]] ✅.

Full Walkthrough10 Steps

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

1🌱 ROOT(Root Seed, Ring 0, d=0)

queue = [3]

queue = [3]. Seed queue with root node 3. Initial output = [].

2🌱 ROOT(Snapshot Ring 0, level_size=1, d=0)

level_size = 1

level_size = 1. Freeze Ring 0 boundary before child enqueue.

3🌱 ROOT(Dequeue 3 & Enqueue Children, Ring 0, d=0)

queue = [9, 20]

Pop 3

\rightarrow

current_level = [3]. Enqueue left child 9 and right child 20

\rightarrow

queue = [9, 20]. Commit Ring 0

\rightarrow

output = [[3]] 📸.

4📸 SNAPSHOT(Snapshot Ring 1, level_size=2, d=1)

queue = [9, 20]

level_size = 2. Freeze Ring 1 boundary. queue = [9, 20].

5⚡ FIFO DRAIN(Dequeue 9, Leaf, Ring 1, d=1)

Pop 9

\rightarrow

current_level = [9]. Node 9 has no children. Remaining in Ring 1: 1 node.

6🌊 ENQUEUE / DISCOVER(Dequeue 20 & Enqueue Children, Ring 1, d=1)

queue = [15, 7]

Pop 20

\rightarrow

current_level = [9, 20]. Enqueue left child 15 and right child 7

\rightarrow

queue = [15, 7]. Commit Ring 1

\rightarrow

output = [[3], [9, 20]] 📸.

7📸 SNAPSHOT(Snapshot Ring 2, level_size=2, d=2)

queue = [15, 7]

level_size = 2. Freeze Ring 2 boundary. queue = [15, 7].

8⚡ FIFO DRAIN(Dequeue 15, Leaf, Ring 2, d=2)

Pop 15

\rightarrow

current_level = [15]. Node 15 has no children. Remaining in Ring 2: 1 node.

9⚡ FIFO DRAIN(Dequeue 7, Leaf, Ring 2, d=2)

Pop 7

\rightarrow

current_level = [15, 7]. Node 7 has no children. Commit Ring 2

\rightarrow

output = [[3], [9, 20], [15, 7]] 📸.

10⚡ FIFO DRAIN(Queue Empty, Complete, d=2)

queue = []

queue = []. All concentric level rings drained. Search Complete: return [[3], [9, 20], [15, 7]] ✅.

Code Shape

Archetype Code Shape

Python

1	queue = deque([root])
2	levels = []
3	while queue:
4	level_size = len(queue)
5	current_level = []
6	for _ in range(level_size):
7	node = queue.popleft()
8	current_level.append(node.val)
9	append the children of node to queue
10	levels.append(current_level)
11	return levels

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

The single line level_size = len(queue), taken before the inner loop starts, is the entire trick behind grouping a flat queue into discrete levels.

🌊 The Analogy: Ripples in a Pond

Drop a stone in a pond and watch the ripples: every point at radius 1 reaches its full circle before any point at radius 2 begins moving. BFS is exactly this, applied to a tree — every node at depth 1 is fully processed before any node at depth 2 is even looked at, because a queue is First-In-First-Out and every depth-1 node was enqueued before any depth-2 node.

🔑 The Key Idea in the Code

Without freezing level_size first, the loop would have no way to know where one ripple ends and the next begins — new nodes keep getting appended to the same queue mid-loop. The snapshot is what lets a single flat queue represent distinct, ordered "rings."

Queue length mutation in loop condition: In languages like Java or C++, querying queue.size() directly inside the loop condition re-evaluates as children are added. Snapshot level_size = len(queue) first.
Memory explosion on wide trees: A full binary tree has $N/2$ nodes at its deepest level. Ensure queue operations are minimal and avoid allocating deep call stacks.
Empty root handling: If root is None, return [] immediately rather than [[]].

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Walk the tree "visiting each depth in left-to-right order" and produce "one list of values per depth": the output is grouped by depth, one list per level. That is Level-Order BFS.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Invariant: at the start of each round, queue holds exactly one level, and level_size is its count. The decision: the round pops exactly level_size nodes, so every child added during it waits for the next round.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

level_size = len(queue) before for _ in range(level_size): the card's trap (a bound that re-reads the queue's length runs into the next level).
if node.left: / if node.right:: push only real children; pushing None makes a later node.val fail.

The 60-Second Interview Pitch

Say this out loud before you type a single line

This is Level-Order BFS. The output groups the values by depth, so I walk the tree with a queue and handle one level per round. At the start of each round I read the queue's length into level size: right then, the queue holds exactly the nodes of one level, left to right. I pop exactly that many, record their values in the current level, and push their left and right children to the back, where they wait for the next round. When the round ends, the queue holds exactly the next level, so the invariant holds again. The trap is re-reading the queue's length while children are being added: the bound keeps growing and the round also processes part of the next level. Freezing it first gives the boundary without storing a depth per node. Every node is pushed and popped once, so it's O(N) time, and the queue holds at most one level plus its children, O(W) space.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

Look at the code: a node enters queue only as root or as a child of a popped node, so each of the N nodes is pushed once and popped once; for _ in range(level_size) over all rounds pops exactly N nodes, each with O(1) work (append its value, test two children). Total O(N).

SPACE COMPLEXITY

O(W) where W is the maximum tree width

queue holds the rest of the current level plus the children already added, never more than about two levels, so O(W). No recursion: the call stack is O(1). The output levels holds all N values and is counted separately.

Formal Recurrence Relation

T(N) = N pushes + N pops = 2N = O(N)

Derivation Progression

Pushes

queue = deque([root]); queue.append(node.left); queue.append(node.right) → N pushes

Each node is pushed once: the root at the start, every other node by its parent.

Pops

for _ in range(level_size): node = queue.popleft() → N pops over all rounds

The rounds' level_size values add up to N, and each pop does O(1) work.

Total

N + N = O(N)

deque.popleft() is O(1), so nothing else grows with N.

Variable Definitions

$N$

Number of nodes in the tree

$W$

Maximum tree width: the most nodes on one level (at most ⌈N/2⌉)

$H$

Height of the tree (used by the brute force)

Memory Architecture & Bounds

🟣 Call Stack

O(1): no recursion

🔵 Auxiliary Heap

O(W): queue holds at most one level plus its children added so far

🟢 Output Space

O(N): levels (not counted as auxiliary)

Boundary Best / Worst Cases

Best Case

$O(N)$

Average Case

$O(N)$

Worst Case

$O(N)$

Graph & Tree Traversal Frontier

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: "one list of values per depth", "left-to-right order": output grouped by depth, Level-Order BFS. At scale the same per-level grouping is what lets results stream out one level at a time instead of after the whole walk.

CONSTRAINTS & BOUNDS

$N \le 2000$ nodes: $2N$ queue operations. Memory is set by the widest level, not the depth: a complete tree of $10^6$ nodes has about $5 \times 10^5$ in its last level, all in queue at once.

FAANG PRODUCTION TRAPS & EDGE CASES

list.pop(0) shifts the whole list: O(K) per pop and O(N²) overall; use collections.deque.popleft(). On a very wide level the queue itself is the memory limit: if levels only need to be counted or summed, keep a running total instead of current_level.

Theory Context•Breadth-First Search (BFS)

MediumLC 102

Binary Tree Level Order Traversal (LeetCode 102)

Target Frequency:AmazonMetaGoogle

Worked Examples

Example 1

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

Output:[[3],[9,20],[15,7]]

Explanation: Depth 0 holds 3, depth 1 holds 9 and 20, depth 2 holds 15 and 7.

Example 2

Input:root = [1]

Output:[[1]]

Explanation: A single node forms one level.

Example 3

Input:root = []

Output:[]

Explanation: An empty tree has no levels.

⚖️Formal Constraints & Bounds

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

Deep-Dive & Conceptual Insights

Core Insight

Taking a snapshot of len(queue) separates nodes of the current level from their newly enqueued children.

Real-World Scenario & Production Applications

Peer-to-peer gossip network broadcasting and organizational hierarchy reporting charts.

Step-by-Step Execution Trace Table

Step-by-Step Wavefront Progression

Step 1 (Root Seed, Ring 0, d=0): queue = [3]. Seed queue with root node 3. Initial output = [].
Step 2 (Snapshot Ring 0, level_size=1, d=0): level_size = 1. Freeze Ring 0 boundary before child enqueue.
Step 3 (Dequeue 3 & Enqueue Children, Ring 0, d=0): Pop 3 $\rightarrow$ current_level = [3]. Enqueue left child 9 and right child 20 $\rightarrow$ queue = [9, 20]. Commit Ring 0 $\rightarrow$ output = [[3]] 📸.
Step 4 (Snapshot Ring 1, level_size=2, d=1): level_size = 2. Freeze Ring 1 boundary. queue = [9, 20].
Step 5 (Dequeue 9, Leaf, Ring 1, d=1): Pop 9 $\rightarrow$ current_level = [9]. Node 9 has no children. Remaining in Ring 1: 1 node.
Step 6 (Dequeue 20 & Enqueue Children, Ring 1, d=1): Pop 20 $\rightarrow$ current_level = [9, 20]. Enqueue left child 15 and right child 7 $\rightarrow$ queue = [15, 7]. Commit Ring 1 $\rightarrow$ output = [[3], [9, 20]] 📸.
Step 7 (Snapshot Ring 2, level_size=2, d=2): level_size = 2. Freeze Ring 2 boundary. queue = [15, 7].
Step 8 (Dequeue 15, Leaf, Ring 2, d=2): Pop 15 $\rightarrow$ current_level = [15]. Node 15 has no children. Remaining in Ring 2: 1 node.
Step 9 (Dequeue 7, Leaf, Ring 2, d=2): Pop 7 $\rightarrow$ current_level = [15, 7]. Node 7 has no children. Commit Ring 2 $\rightarrow$ output = [[3], [9, 20], [15, 7]] 📸.
Step 10 (Queue Empty, Complete, d=2): queue = []. All concentric level rings drained. Search Complete: return [[3], [9, 20], [15, 7]] ✅.

Full Walkthrough10 Steps

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

1🌱 ROOT(Root Seed, Ring 0, d=0)

queue = [3]

queue = [3]. Seed queue with root node 3. Initial output = [].

2🌱 ROOT(Snapshot Ring 0, level_size=1, d=0)

level_size = 1

level_size = 1. Freeze Ring 0 boundary before child enqueue.

3🌱 ROOT(Dequeue 3 & Enqueue Children, Ring 0, d=0)

queue = [9, 20]

Pop 3

\rightarrow

current_level = [3]. Enqueue left child 9 and right child 20

\rightarrow

queue = [9, 20]. Commit Ring 0

\rightarrow

output = [[3]] 📸.

4📸 SNAPSHOT(Snapshot Ring 1, level_size=2, d=1)

queue = [9, 20]

level_size = 2. Freeze Ring 1 boundary. queue = [9, 20].

5⚡ FIFO DRAIN(Dequeue 9, Leaf, Ring 1, d=1)

Pop 9

\rightarrow

current_level = [9]. Node 9 has no children. Remaining in Ring 1: 1 node.

6🌊 ENQUEUE / DISCOVER(Dequeue 20 & Enqueue Children, Ring 1, d=1)

queue = [15, 7]

Pop 20

\rightarrow

current_level = [9, 20]. Enqueue left child 15 and right child 7

\rightarrow

queue = [15, 7]. Commit Ring 1

\rightarrow

output = [[3], [9, 20]] 📸.

7📸 SNAPSHOT(Snapshot Ring 2, level_size=2, d=2)

queue = [15, 7]

level_size = 2. Freeze Ring 2 boundary. queue = [15, 7].

8⚡ FIFO DRAIN(Dequeue 15, Leaf, Ring 2, d=2)

Pop 15

\rightarrow

current_level = [15]. Node 15 has no children. Remaining in Ring 2: 1 node.

9⚡ FIFO DRAIN(Dequeue 7, Leaf, Ring 2, d=2)

Pop 7

\rightarrow

current_level = [15, 7]. Node 7 has no children. Commit Ring 2

\rightarrow

output = [[3], [9, 20], [15, 7]] 📸.

10⚡ FIFO DRAIN(Queue Empty, Complete, d=2)

queue = []

queue = []. All concentric level rings drained. Search Complete: return [[3], [9, 20], [15, 7]] ✅.

Code Shape

Archetype Code Shape

Python

1	queue = deque([root])
2	levels = []
3	while queue:
4	level_size = len(queue)
5	current_level = []
6	for _ in range(level_size):
7	node = queue.popleft()
8	current_level.append(node.val)
9	append the children of node to queue
10	levels.append(current_level)
11	return levels

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

The single line level_size = len(queue), taken before the inner loop starts, is the entire trick behind grouping a flat queue into discrete levels.

🌊 The Analogy: Ripples in a Pond

🔑 The Key Idea in the Code

Queue length mutation in loop condition: In languages like Java or C++, querying queue.size() directly inside the loop condition re-evaluates as children are added. Snapshot level_size = len(queue) first.
Memory explosion on wide trees: A full binary tree has $N/2$ nodes at its deepest level. Ensure queue operations are minimal and avoid allocating deep call stacks.
Empty root handling: If root is None, return [] immediately rather than [[]].

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Walk the tree "visiting each depth in left-to-right order" and produce "one list of values per depth": the output is grouped by depth, one list per level. That is Level-Order BFS.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

level_size = len(queue) before for _ in range(level_size): the card's trap (a bound that re-reads the queue's length runs into the next level).
if node.left: / if node.right:: push only real children; pushing None makes a later node.val fail.

The 60-Second Interview Pitch

Say this out loud before you type a single line

This is Level-Order BFS. The output groups the values by depth, so I walk the tree with a queue and handle one level per round. At the start of each round I read the queue's length into level size: right then, the queue holds exactly the nodes of one level, left to right. I pop exactly that many, record their values in the current level, and push their left and right children to the back, where they wait for the next round. When the round ends, the queue holds exactly the next level, so the invariant holds again. The trap is re-reading the queue's length while children are being added: the bound keeps growing and the round also processes part of the next level. Freezing it first gives the boundary without storing a depth per node. Every node is pushed and popped once, so it's O(N) time, and the queue holds at most one level plus its children, O(W) space.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

SPACE COMPLEXITY

O(W) where W is the maximum tree width

Formal Recurrence Relation

T(N) = N pushes + N pops = 2N = O(N)

Derivation Progression

Pushes

queue = deque([root]); queue.append(node.left); queue.append(node.right) → N pushes

Each node is pushed once: the root at the start, every other node by its parent.

Pops

for _ in range(level_size): node = queue.popleft() → N pops over all rounds

The rounds' level_size values add up to N, and each pop does O(1) work.

Total

N + N = O(N)

deque.popleft() is O(1), so nothing else grows with N.

Variable Definitions

$N$

Number of nodes in the tree

$W$

Maximum tree width: the most nodes on one level (at most ⌈N/2⌉)

$H$

Height of the tree (used by the brute force)

Memory Architecture & Bounds

🟣 Call Stack

O(1): no recursion

🔵 Auxiliary Heap

O(W): queue holds at most one level plus its children added so far

🟢 Output Space

O(N): levels (not counted as auxiliary)

Boundary Best / Worst Cases

Best Case

$O(N)$

Average Case

$O(N)$

Worst Case

$O(N)$

Graph & Tree Traversal Frontier

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

CONSTRAINTS & BOUNDS

$N \le 2000$ nodes: $2N$ queue operations. Memory is set by the widest level, not the depth: a complete tree of $10^6$ nodes has about $5 \times 10^5$ in its last level, all in queue at once.

FAANG PRODUCTION TRAPS & EDGE CASES

Theory Context•Breadth-First Search (BFS)

MediumLC 102

Binary Tree Level Order Traversal (LeetCode 102)

Target Frequency:AmazonMetaGoogle

Worked Examples

Example 1

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

Output:[[3],[9,20],[15,7]]

Explanation: Depth 0 holds 3, depth 1 holds 9 and 20, depth 2 holds 15 and 7.

Example 2

Input:root = [1]

Output:[[1]]

Explanation: A single node forms one level.

Example 3

Input:root = []

Output:[]

Explanation: An empty tree has no levels.

⚖️Formal Constraints & Bounds

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

Deep-Dive & Conceptual Insights

Core Insight

Taking a snapshot of len(queue) separates nodes of the current level from their newly enqueued children.

Real-World Scenario & Production Applications

Peer-to-peer gossip network broadcasting and organizational hierarchy reporting charts.

Step-by-Step Execution Trace Table

Step-by-Step Wavefront Progression

Step 1 (Root Seed, Ring 0, d=0): queue = [3]. Seed queue with root node 3. Initial output = [].
Step 2 (Snapshot Ring 0, level_size=1, d=0): level_size = 1. Freeze Ring 0 boundary before child enqueue.
Step 3 (Dequeue 3 & Enqueue Children, Ring 0, d=0): Pop 3 $\rightarrow$ current_level = [3]. Enqueue left child 9 and right child 20 $\rightarrow$ queue = [9, 20]. Commit Ring 0 $\rightarrow$ output = [[3]] 📸.
Step 4 (Snapshot Ring 1, level_size=2, d=1): level_size = 2. Freeze Ring 1 boundary. queue = [9, 20].
Step 5 (Dequeue 9, Leaf, Ring 1, d=1): Pop 9 $\rightarrow$ current_level = [9]. Node 9 has no children. Remaining in Ring 1: 1 node.
Step 6 (Dequeue 20 & Enqueue Children, Ring 1, d=1): Pop 20 $\rightarrow$ current_level = [9, 20]. Enqueue left child 15 and right child 7 $\rightarrow$ queue = [15, 7]. Commit Ring 1 $\rightarrow$ output = [[3], [9, 20]] 📸.
Step 7 (Snapshot Ring 2, level_size=2, d=2): level_size = 2. Freeze Ring 2 boundary. queue = [15, 7].
Step 8 (Dequeue 15, Leaf, Ring 2, d=2): Pop 15 $\rightarrow$ current_level = [15]. Node 15 has no children. Remaining in Ring 2: 1 node.
Step 9 (Dequeue 7, Leaf, Ring 2, d=2): Pop 7 $\rightarrow$ current_level = [15, 7]. Node 7 has no children. Commit Ring 2 $\rightarrow$ output = [[3], [9, 20], [15, 7]] 📸.
Step 10 (Queue Empty, Complete, d=2): queue = []. All concentric level rings drained. Search Complete: return [[3], [9, 20], [15, 7]] ✅.

Full Walkthrough10 Steps

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

1🌱 ROOT(Root Seed, Ring 0, d=0)

queue = [3]

queue = [3]. Seed queue with root node 3. Initial output = [].

2🌱 ROOT(Snapshot Ring 0, level_size=1, d=0)

level_size = 1

level_size = 1. Freeze Ring 0 boundary before child enqueue.

3🌱 ROOT(Dequeue 3 & Enqueue Children, Ring 0, d=0)

queue = [9, 20]

Pop 3

\rightarrow

current_level = [3]. Enqueue left child 9 and right child 20

\rightarrow

queue = [9, 20]. Commit Ring 0

\rightarrow

output = [[3]] 📸.

4📸 SNAPSHOT(Snapshot Ring 1, level_size=2, d=1)

queue = [9, 20]

level_size = 2. Freeze Ring 1 boundary. queue = [9, 20].

5⚡ FIFO DRAIN(Dequeue 9, Leaf, Ring 1, d=1)

Pop 9

\rightarrow

current_level = [9]. Node 9 has no children. Remaining in Ring 1: 1 node.

6🌊 ENQUEUE / DISCOVER(Dequeue 20 & Enqueue Children, Ring 1, d=1)

queue = [15, 7]

Pop 20

\rightarrow

current_level = [9, 20]. Enqueue left child 15 and right child 7

\rightarrow

queue = [15, 7]. Commit Ring 1

\rightarrow

output = [[3], [9, 20]] 📸.

7📸 SNAPSHOT(Snapshot Ring 2, level_size=2, d=2)

queue = [15, 7]

level_size = 2. Freeze Ring 2 boundary. queue = [15, 7].

8⚡ FIFO DRAIN(Dequeue 15, Leaf, Ring 2, d=2)

Pop 15

\rightarrow

current_level = [15]. Node 15 has no children. Remaining in Ring 2: 1 node.

9⚡ FIFO DRAIN(Dequeue 7, Leaf, Ring 2, d=2)

Pop 7

\rightarrow

current_level = [15, 7]. Node 7 has no children. Commit Ring 2

\rightarrow

output = [[3], [9, 20], [15, 7]] 📸.

10⚡ FIFO DRAIN(Queue Empty, Complete, d=2)

queue = []

queue = []. All concentric level rings drained. Search Complete: return [[3], [9, 20], [15, 7]] ✅.

Code Shape

Archetype Code Shape

Python

1	queue = deque([root])
2	levels = []
3	while queue:
4	level_size = len(queue)
5	current_level = []
6	for _ in range(level_size):
7	node = queue.popleft()
8	current_level.append(node.val)
9	append the children of node to queue
10	levels.append(current_level)
11	return levels

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

The single line level_size = len(queue), taken before the inner loop starts, is the entire trick behind grouping a flat queue into discrete levels.

🌊 The Analogy: Ripples in a Pond

🔑 The Key Idea in the Code

Queue length mutation in loop condition: In languages like Java or C++, querying queue.size() directly inside the loop condition re-evaluates as children are added. Snapshot level_size = len(queue) first.
Memory explosion on wide trees: A full binary tree has $N/2$ nodes at its deepest level. Ensure queue operations are minimal and avoid allocating deep call stacks.
Empty root handling: If root is None, return [] immediately rather than [[]].

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Walk the tree "visiting each depth in left-to-right order" and produce "one list of values per depth": the output is grouped by depth, one list per level. That is Level-Order BFS.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

level_size = len(queue) before for _ in range(level_size): the card's trap (a bound that re-reads the queue's length runs into the next level).
if node.left: / if node.right:: push only real children; pushing None makes a later node.val fail.

The 60-Second Interview Pitch

Say this out loud before you type a single line

This is Level-Order BFS. The output groups the values by depth, so I walk the tree with a queue and handle one level per round. At the start of each round I read the queue's length into level size: right then, the queue holds exactly the nodes of one level, left to right. I pop exactly that many, record their values in the current level, and push their left and right children to the back, where they wait for the next round. When the round ends, the queue holds exactly the next level, so the invariant holds again. The trap is re-reading the queue's length while children are being added: the bound keeps growing and the round also processes part of the next level. Freezing it first gives the boundary without storing a depth per node. Every node is pushed and popped once, so it's O(N) time, and the queue holds at most one level plus its children, O(W) space.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

SPACE COMPLEXITY

O(W) where W is the maximum tree width

Formal Recurrence Relation

T(N) = N pushes + N pops = 2N = O(N)

Derivation Progression

Pushes

queue = deque([root]); queue.append(node.left); queue.append(node.right) → N pushes

Each node is pushed once: the root at the start, every other node by its parent.

Pops

for _ in range(level_size): node = queue.popleft() → N pops over all rounds

The rounds' level_size values add up to N, and each pop does O(1) work.

Total

N + N = O(N)

deque.popleft() is O(1), so nothing else grows with N.

Variable Definitions

$N$

Number of nodes in the tree

$W$

Maximum tree width: the most nodes on one level (at most ⌈N/2⌉)

$H$

Height of the tree (used by the brute force)

Memory Architecture & Bounds

🟣 Call Stack

O(1): no recursion

🔵 Auxiliary Heap

O(W): queue holds at most one level plus its children added so far

🟢 Output Space

O(N): levels (not counted as auxiliary)

Boundary Best / Worst Cases

Best Case

$O(N)$

Average Case

$O(N)$

Worst Case

$O(N)$

Graph & Tree Traversal Frontier

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

CONSTRAINTS & BOUNDS

$N \le 2000$ nodes: $2N$ queue operations. Memory is set by the widest level, not the depth: a complete tree of $10^6$ nodes has about $5 \times 10^5$ in its last level, all in queue at once.

FAANG PRODUCTION TRAPS & EDGE CASES

Theory Context•Breadth-First Search (BFS)

MediumLC 102

Binary Tree Level Order Traversal (LeetCode 102)

Target Frequency:AmazonMetaGoogle

Worked Examples

Example 1

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

Output:[[3],[9,20],[15,7]]

Explanation: Depth 0 holds 3, depth 1 holds 9 and 20, depth 2 holds 15 and 7.

Example 2

Input:root = [1]

Output:[[1]]

Explanation: A single node forms one level.

Example 3

Input:root = []

Output:[]

Explanation: An empty tree has no levels.

⚖️Formal Constraints & Bounds

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

Deep-Dive & Conceptual Insights

Core Insight

Taking a snapshot of len(queue) separates nodes of the current level from their newly enqueued children.

Real-World Scenario & Production Applications

Peer-to-peer gossip network broadcasting and organizational hierarchy reporting charts.

Step-by-Step Execution Trace Table

Step-by-Step Wavefront Progression

Step 1 (Root Seed, Ring 0, d=0): queue = [3]. Seed queue with root node 3. Initial output = [].
Step 2 (Snapshot Ring 0, level_size=1, d=0): level_size = 1. Freeze Ring 0 boundary before child enqueue.
Step 3 (Dequeue 3 & Enqueue Children, Ring 0, d=0): Pop 3 $\rightarrow$ current_level = [3]. Enqueue left child 9 and right child 20 $\rightarrow$ queue = [9, 20]. Commit Ring 0 $\rightarrow$ output = [[3]] 📸.
Step 4 (Snapshot Ring 1, level_size=2, d=1): level_size = 2. Freeze Ring 1 boundary. queue = [9, 20].
Step 5 (Dequeue 9, Leaf, Ring 1, d=1): Pop 9 $\rightarrow$ current_level = [9]. Node 9 has no children. Remaining in Ring 1: 1 node.
Step 6 (Dequeue 20 & Enqueue Children, Ring 1, d=1): Pop 20 $\rightarrow$ current_level = [9, 20]. Enqueue left child 15 and right child 7 $\rightarrow$ queue = [15, 7]. Commit Ring 1 $\rightarrow$ output = [[3], [9, 20]] 📸.
Step 7 (Snapshot Ring 2, level_size=2, d=2): level_size = 2. Freeze Ring 2 boundary. queue = [15, 7].
Step 8 (Dequeue 15, Leaf, Ring 2, d=2): Pop 15 $\rightarrow$ current_level = [15]. Node 15 has no children. Remaining in Ring 2: 1 node.
Step 9 (Dequeue 7, Leaf, Ring 2, d=2): Pop 7 $\rightarrow$ current_level = [15, 7]. Node 7 has no children. Commit Ring 2 $\rightarrow$ output = [[3], [9, 20], [15, 7]] 📸.
Step 10 (Queue Empty, Complete, d=2): queue = []. All concentric level rings drained. Search Complete: return [[3], [9, 20], [15, 7]] ✅.

Full Walkthrough10 Steps

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

1🌱 ROOT(Root Seed, Ring 0, d=0)

queue = [3]

queue = [3]. Seed queue with root node 3. Initial output = [].

2🌱 ROOT(Snapshot Ring 0, level_size=1, d=0)

level_size = 1

level_size = 1. Freeze Ring 0 boundary before child enqueue.

3🌱 ROOT(Dequeue 3 & Enqueue Children, Ring 0, d=0)

queue = [9, 20]

Pop 3

\rightarrow

current_level = [3]. Enqueue left child 9 and right child 20

\rightarrow

queue = [9, 20]. Commit Ring 0

\rightarrow

output = [[3]] 📸.

4📸 SNAPSHOT(Snapshot Ring 1, level_size=2, d=1)

queue = [9, 20]

level_size = 2. Freeze Ring 1 boundary. queue = [9, 20].

5⚡ FIFO DRAIN(Dequeue 9, Leaf, Ring 1, d=1)

Pop 9

\rightarrow

current_level = [9]. Node 9 has no children. Remaining in Ring 1: 1 node.

6🌊 ENQUEUE / DISCOVER(Dequeue 20 & Enqueue Children, Ring 1, d=1)

queue = [15, 7]

Pop 20

\rightarrow

current_level = [9, 20]. Enqueue left child 15 and right child 7

\rightarrow

queue = [15, 7]. Commit Ring 1

\rightarrow

output = [[3], [9, 20]] 📸.

7📸 SNAPSHOT(Snapshot Ring 2, level_size=2, d=2)

queue = [15, 7]

level_size = 2. Freeze Ring 2 boundary. queue = [15, 7].

8⚡ FIFO DRAIN(Dequeue 15, Leaf, Ring 2, d=2)

Pop 15

\rightarrow

current_level = [15]. Node 15 has no children. Remaining in Ring 2: 1 node.

9⚡ FIFO DRAIN(Dequeue 7, Leaf, Ring 2, d=2)

Pop 7

\rightarrow

current_level = [15, 7]. Node 7 has no children. Commit Ring 2

\rightarrow

output = [[3], [9, 20], [15, 7]] 📸.

10⚡ FIFO DRAIN(Queue Empty, Complete, d=2)

queue = []

queue = []. All concentric level rings drained. Search Complete: return [[3], [9, 20], [15, 7]] ✅.

Code Shape

Archetype Code Shape

Python

1	queue = deque([root])
2	levels = []
3	while queue:
4	level_size = len(queue)
5	current_level = []
6	for _ in range(level_size):
7	node = queue.popleft()
8	current_level.append(node.val)
9	append the children of node to queue
10	levels.append(current_level)
11	return levels

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

The single line level_size = len(queue), taken before the inner loop starts, is the entire trick behind grouping a flat queue into discrete levels.

🌊 The Analogy: Ripples in a Pond

🔑 The Key Idea in the Code

Queue length mutation in loop condition: In languages like Java or C++, querying queue.size() directly inside the loop condition re-evaluates as children are added. Snapshot level_size = len(queue) first.
Memory explosion on wide trees: A full binary tree has $N/2$ nodes at its deepest level. Ensure queue operations are minimal and avoid allocating deep call stacks.
Empty root handling: If root is None, return [] immediately rather than [[]].

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Walk the tree "visiting each depth in left-to-right order" and produce "one list of values per depth": the output is grouped by depth, one list per level. That is Level-Order BFS.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

level_size = len(queue) before for _ in range(level_size): the card's trap (a bound that re-reads the queue's length runs into the next level).
if node.left: / if node.right:: push only real children; pushing None makes a later node.val fail.

The 60-Second Interview Pitch

Say this out loud before you type a single line

This is Level-Order BFS. The output groups the values by depth, so I walk the tree with a queue and handle one level per round. At the start of each round I read the queue's length into level size: right then, the queue holds exactly the nodes of one level, left to right. I pop exactly that many, record their values in the current level, and push their left and right children to the back, where they wait for the next round. When the round ends, the queue holds exactly the next level, so the invariant holds again. The trap is re-reading the queue's length while children are being added: the bound keeps growing and the round also processes part of the next level. Freezing it first gives the boundary without storing a depth per node. Every node is pushed and popped once, so it's O(N) time, and the queue holds at most one level plus its children, O(W) space.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

SPACE COMPLEXITY

O(W) where W is the maximum tree width

Formal Recurrence Relation

T(N) = N pushes + N pops = 2N = O(N)

Derivation Progression

Pushes

queue = deque([root]); queue.append(node.left); queue.append(node.right) → N pushes

Each node is pushed once: the root at the start, every other node by its parent.

Pops

for _ in range(level_size): node = queue.popleft() → N pops over all rounds

The rounds' level_size values add up to N, and each pop does O(1) work.

Total

N + N = O(N)

deque.popleft() is O(1), so nothing else grows with N.

Variable Definitions

$N$

Number of nodes in the tree

$W$

Maximum tree width: the most nodes on one level (at most ⌈N/2⌉)

$H$

Height of the tree (used by the brute force)

Memory Architecture & Bounds

🟣 Call Stack

O(1): no recursion

🔵 Auxiliary Heap

O(W): queue holds at most one level plus its children added so far

🟢 Output Space

O(N): levels (not counted as auxiliary)

Boundary Best / Worst Cases

Best Case

$O(N)$

Average Case

$O(N)$

Worst Case

$O(N)$

Graph & Tree Traversal Frontier

Synthesizing vector architecture diagram...

Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

CONSTRAINTS & BOUNDS

$N \le 2000$ nodes: $2N$ queue operations. Memory is set by the widest level, not the depth: a complete tree of $10^6$ nodes has about $5 \times 10^5$ in its last level, all in queue at once.

FAANG PRODUCTION TRAPS & EDGE CASES