N-ary Tree Level Order Traversal

📋

Problem

Imagine you are organizing a company hierarchy meeting and want to invite employees level by level, starting from the CEO down to the lowest level employees.

Given an n-ary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level). The input is the root of the n-ary tree. Output a list of lists, where each inner list contains the values of nodes at that level.

The number of nodes in the tree is in the range [1, 10^5].0 <= Node.val <= 10^5The height of the n-ary tree is less than or equal to 1000.

Edge cases: Empty tree (root = null) → []Tree with only one node → [[root.val]]Tree where nodes have varying number of children → correctly group by levels

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IDE

def levelOrder(root: 'Node') -> list:public List<List<Integer>> levelOrder(Node root)vector<vector<int>> levelOrder(Node* root)function levelOrder(root)

def levelOrder(root):
    # Write your solution here
    pass

class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        // Write your solution here
        return new ArrayList<>();
    }
}

#include <vector>
using namespace std;

vector<vector<int>> levelOrder(Node* root) {
    // Write your solution here
    return {};
}

function levelOrder(root) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: [[1,3,2,4],[5,6]]Incorrect BFS grouping: all nodes processed in one level instead of separate levels.✅ Use queue size to separate nodes by level and group them accordingly.

Wrong: [[1],[3,2,4],[5]]Missing some children in last level due to incomplete children iteration.✅ Iterate over all children of each node when enqueuing for next level.

Wrong: [[1],[2,3,4,5]]Fails on empty tree input or null root, causing crash or wrong output.✅ Add null root check at start and return empty list if root is null.

Wrong: [[1],[2,3,4],[5,6,7,8]]Off-by-one error causing extra nodes or levels merged incorrectly.✅ Ensure level index increments correctly and nodes are grouped per exact level.

Wrong: Timeout or stack overflowRecursive solution without tail recursion or iterative BFS causing stack overflow or TLE on large input.✅ Use iterative BFS with queue to achieve O(n) time and avoid recursion depth limits.

✓

Test Cases

Make sure your code handles empty and single-node trees correctly.

Test trees with varying children counts and deep chains to catch subtle bugs.

Optimize your BFS to handle large inputs efficiently within time limits.

t1_01basic

Input{"root":[1]}

Expected[[1],[3,2,4],[5,6]]

⏱ Performance - must finish in 2000ms

Level 0 has node 1; Level 1 has nodes 3, 2, 4; Level 2 has nodes 5, 6.

💡 Think about traversing the tree level by level using a queue.

💡 Use a queue to process nodes in FIFO order and group nodes by levels.

💡 Enqueue children of current level nodes before moving to the next level.

Why it failed: Output differs from expected level grouping; likely incorrect BFS or level tracking. Fix by ensuring nodes are grouped by levels using a queue and processing all nodes at each level before moving on.

✓ Correct BFS level order traversal implemented.

t1_02basic

Input{"root":[10]}

Expected[[10],[20,30],[40,50,60]]

⏱ Performance - must finish in 2000ms

Level 0: 10; Level 1: 20, 30; Level 2: 40, 50, 60.

💡 Use a queue to process nodes level by level.

💡 At each iteration, process all nodes currently in the queue to form one level.

💡 Add children of all nodes in current level to the queue for next iteration.

Why it failed: Incorrect grouping of nodes by level; possibly mixing nodes from different levels. Fix by processing nodes level-wise using queue size to separate levels.

✓ Correctly groups nodes by level using BFS.

t2_01edge

Input{"root":null}

Expected[]

⏱ Performance - must finish in 2000ms

Empty tree returns empty list.

💡 Check if root is null before processing.

💡 Return empty list immediately if tree is empty.

💡 Add a base case to handle null root input.

Why it failed: Function crashes or returns non-empty output on null root. Fix by adding a null check at start and returning [].

✓ Handles empty tree input correctly.

t2_02edge

Input{"root":[42]}

Expected[[42]]

⏱ Performance - must finish in 2000ms

Single node tree returns list with one level containing the root value.

💡 Test with tree having only one node and no children.

💡 Ensure the root node's value is included in the output.

💡 Return [[root.val]] when root has no children.

Why it failed: Output missing root node or empty list returned for single node tree. Fix by always adding root value as first level.

✓ Correctly returns single node level order.

t2_03edge

Input{"root":[1]}

Expected[[1],[2],[3],[4]]

⏱ Performance - must finish in 2000ms

Deep tree with single child chains outputs each node in its own level list.

💡 Consider trees where each node has exactly one child.

💡 Ensure level order traversal handles deep chains correctly.

💡 Each recursive or iterative step should add one level list per node depth.

Why it failed: Output merges multiple levels or misses levels in deep single-child chains. Fix by correctly incrementing level and grouping nodes per depth.

✓ Correctly handles deep single-child chains.

t3_01corner

Input{"root":[1]}

Expected[[1],[2,3,4,5]]

⏱ Performance - must finish in 2000ms

Wide tree with many children at one level; all children grouped correctly.

💡 Check that all children of a node are enqueued before moving to next level.

💡 Avoid stopping traversal early or processing only some children.

💡 Ensure BFS processes all nodes in current level before next.

Why it failed: Output missing some children or partial level output. Fix by enqueuing all children of current level nodes before next iteration.

✓ Correctly processes wide levels with many children.

t3_02corner

Input{"root":[1]}

Expected[[1],[2,3,4],[5,6,7]]

⏱ Performance - must finish in 2000ms

Tree with varying number of children per node; levels grouped correctly.

💡 Test trees where nodes have different numbers of children.

💡 Ensure BFS does not assume fixed number of children per node.

💡 Process children list dynamically for each node.

Why it failed: Output assumes fixed children count or misses some children. Fix by iterating over each node's children list fully.

✓ Correctly handles nodes with varying children counts.

t3_03corner

Input{"root":[1]}

Expected[[1],[2],[3],[4],[5]]

⏱ Performance - must finish in 2000ms

Deep chain tests off-by-one errors in level counting.

💡 Check that level increments correctly at each recursion or iteration.

💡 Avoid off-by-one errors in indexing or level tracking.

💡 Verify that each node is placed in the correct level list.

Why it failed: Levels merged or skipped due to off-by-one error in level index. Fix by ensuring level index increments exactly once per depth.

✓ Correctly tracks levels without off-by-one errors.

t4_01performance

Input{"root":[1]}

Expectednull

⏱ Performance - must finish in 2000ms

Large tree with 100000 nodes in a balanced n-ary structure; BFS must complete in 2s.

💡 Use iterative BFS with queue for O(n) time complexity.

💡 Avoid recursion depth limits by iterative approach.

💡 Process nodes level by level efficiently without extra overhead.

Why it failed: Solution times out due to exponential or inefficient traversal. Fix by using O(n) BFS with queue instead of recursion or repeated scanning.

✓ Solution runs within time limits using O(n) BFS.

Practice

1. You need to collect nodes of a binary tree grouped by their depth level, from top to bottom, left to right. Which algorithmic approach guarantees an optimal solution that visits each node exactly once and groups nodes by their level?

easy

A. Use a Depth-First Search (DFS) without tracking levels, then sort nodes by depth after traversal.

B. Use a recursive approach that collects nodes in preorder traversal order.

C. Use Breadth-First Search (BFS) with a queue to process nodes level by level.

D. Use a greedy approach that picks the leftmost node at each level before moving to the next.

Solution

1. Step 1: Understand the problem requirement

   The problem requires grouping nodes by their depth level, visiting nodes level by level from top to bottom and left to right.

2. Step 2: Evaluate algorithm suitability

   BFS with a queue naturally processes nodes level by level, ensuring each node is visited once and grouped correctly. DFS without level tracking or greedy approaches do not guarantee correct grouping or order.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   BFS processes nodes in level order by design [OK]

Hint: BFS naturally processes nodes level by level [OK]

Common Mistakes:

• Thinking DFS without level tracking groups nodes by level
• Assuming greedy leftmost picks produce level order
• Sorting after DFS is inefficient

2. You need to find the value of the leftmost node at the deepest level of a binary tree. Which approach guarantees an optimal solution that efficiently finds this value without traversing unnecessary nodes multiple times?

easy

A. Breadth-first search (BFS) level order traversal inserting right child before left child to ensure the last node processed is the bottom-left node.

B. Depth-first search (DFS) that tracks the maximum depth and updates the leftmost node when a deeper level is found.

C. Greedy traversal always choosing the left child first until reaching the deepest node.

D. Dynamic programming storing results of subtrees to combine leftmost values at each depth.

Solution

1. Step 1: Understand problem goal

   The problem requires finding the bottom-left value, which is the leftmost node at the deepest level.

2. Step 2: Analyze BFS approach with right-to-left insertion

   BFS with queue inserting right child before left child ensures the last node popped is the bottom-left node, but this approach requires careful enqueue order and may be less intuitive.

3. Step 3: Analyze DFS approach

   DFS that tracks maximum depth and updates the leftmost node when a deeper level is found directly finds the bottom-left node efficiently without relying on enqueue order.

4. Step 4: Choose optimal approach

   DFS approach is simpler and guarantees correct bottom-left node without tricky enqueue order.

5. Final Answer:

   Option B -> Option B

6. Quick Check:

   DFS tracking max depth and leftmost node is a standard, efficient solution [OK]

Hint: DFS tracking max depth and leftmost node efficiently finds bottom-left node [OK]

Common Mistakes:

• Assuming BFS with right-to-left insertion is always simpler
• Thinking greedy left child traversal always finds bottom-left
• Misapplying DP to this problem

3. You are given a binary tree and need to find the maximum width among all levels. The width is defined as the length between the end-nodes (the leftmost and rightmost non-null nodes), including null nodes in between. Which approach guarantees an optimal solution for this problem?

easy

A. Use BFS with indexing of nodes to track their positions and normalize indices at each level to avoid overflow.

B. Perform a simple level order traversal counting only non-null nodes at each level and take the maximum count.

C. Use DFS to traverse the tree and track the minimum and maximum node indices at each depth to compute width.

D. Apply a greedy approach that always picks the widest subtree at each node to estimate maximum width.

Solution

1. Step 1: Understand the problem constraints

   The width includes null nodes between end-nodes, so counting only non-null nodes per level underestimates width.

2. Step 2: Identify the approach that tracks positions

   BFS with indexing assigns position indices to nodes, allowing calculation of width as the difference between first and last indices at each level. Normalizing indices prevents integer overflow on deep trees.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Indexing with normalization handles sparse trees and deep levels correctly [OK]

Hint: Index nodes with BFS and normalize indices [OK]

Common Mistakes:

• Counting only non-null nodes per level
• Using DFS without proper indexing
• Ignoring index normalization causing overflow

4. What is the space complexity of the optimal BFS approach for computing the average of levels in a binary tree, where n is the number of nodes and w is the maximum width of the tree?

medium

A. O(log n) because the tree height bounds the queue size

B. O(n) because all nodes are stored in the queue at once

C. O(w) because the queue holds at most all nodes at the widest level

D. O(n) due to storing all node values in the result list

Solution

1. Step 1: Identify maximum queue size during BFS

   The queue holds nodes level-by-level; max size is the maximum number of nodes at any level, i.e., width w.

2. Step 2: Consider auxiliary space for result list

   Result list stores one average per level, at most height h, which is ≤ n but smaller than w in wide trees.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Queue size dominates space, bounded by maximum width w [OK]

Hint: Queue size = max nodes at any level = width w [OK]

Common Mistakes:

• Assuming queue holds all nodes simultaneously (O(n))
• Confusing height with width for queue size
• Ignoring space for result list

5. What is the time complexity of the optimal BFS approach with indexing and normalization to find the maximum width of a binary tree with n nodes?

medium

A. O(n log n) because of sorting indices at each level

B. O(n) because each node is visited exactly once in BFS

C. O(n^2) due to repeated index normalization at each level

D. O(n) but with extra O(n) space for the queue and index tracking

Solution

1. Step 1: Analyze BFS traversal

   Each node is enqueued and dequeued exactly once, so BFS visits all n nodes in O(n) time.

2. Step 2: Consider index normalization cost

   Index normalization is a constant-time arithmetic operation per node, not sorting or scanning the entire level.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Linear time traversal with constant work per node [OK]

Hint: BFS visits each node once, normalization is O(1) per node [OK]

Common Mistakes:

• Assuming sorting indices per level adds log factor
• Confusing normalization with expensive operations
• Mistaking queue size for nested loops causing O(n^2)