Level Order Traversal — Problem Statement & Solution Guide
Problem Description
Given the root of a binary tree, determine the sequence of node values encountered at each depth level, starting from the root (level 0) and proceeding downwards. The traversal must process nodes from left to right within each level. Return a collection of lists, where the i-th list contains the values of all nodes at depth i, ordered by their horizontal position from left to right.
The input is provided as the root node of a binary tree. Each node contains an integer value and pointers to its left and right children. If a child pointer is null, that branch is considered empty. The output should be a 2D array (or list of lists) where each inner array corresponds to a specific level of the tree. If the tree is empty (root is null), return an empty array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Level Order Traversal"
WHY DOES IT MATTER?
Breadth‑first traversal is a core graph‑theoretic pattern that guarantees minimal distance ordering, which is essential for problems that require processing nodes level by level, such as finding the shortest path in unweighted graphs, serializing trees, or generating hierarchical UI layouts.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that a single queue can implicitly track level boundaries by noting the queue size at the start of each iteration, eliminating the need for nested loops or repeated tree scans, thus collapsing the time complexity to O(n).
REAL-WORLD CONNECTION
Think of a multi‑storey building evacuation: the fire marshal directs people floor by floor, starting from the ground level and moving upward, ensuring everyone on a given floor exits before the next floor is addressed—mirroring BFS’s level‑wise processing of nodes.
During the interview, initialize the queue with the root, then use a while‑loop that captures the current queue length as the level size; this pattern makes the code concise, avoids off‑by‑one errors, and clearly communicates the level separation to the interviewer.
COMPLEXITY AT A GLANCE
O(n)O(w) where w is the maximum width of the tree (worst‑case O(n))Core Theory — Why This Approach?
Level order traversal, also known as breadth‑first search (BFS) on a binary tree, visits nodes layer by layer starting at the root and moving outward. The natural way to achieve this ordering is to use a first‑in‑first‑out (FIFO) queue: enqueue the root, then repeatedly dequeue a node, record its value, and enqueue its left and right children. This guarantees that all nodes at depth *d* are processed before any node at depth *d+1*, preserving the left‑to‑right horizontal order because children are enqueued in that exact sequence. Naïve recursive approaches that mimic depth‑first traversals (pre‑order, in‑order, post‑order) cannot directly produce level‑wise grouping without extra bookkeeping, leading to O(n²) time when repeatedly scanning the tree for each depth. The optimal BFS paradigm runs in linear time O(n) because each node is visited exactly once, and it uses O(w) auxiliary space where *w* is the maximum width of the tree (the number of nodes at the widest level), which is bounded by O(n) in the worst case but typically far smaller. This pattern scales to massive trees and is the foundation for many higher‑level algorithms such as shortest‑path in unweighted graphs, serialization of trees, and parallel processing of hierarchical data.
Interview Questions on This Problem
Q1How would you modify the standard BFS level order traversal to return the nodes of each level in reverse order (right to left) while still using O(n) time?
Use a deque instead of a simple queue: for each level, iterate over the current size, pop nodes from the front, and push their children onto the back in left‑to‑right order, but prepend each popped node's value to the level list (or alternatively, push children onto the front in right‑to‑left order). This yields a right‑to‑left ordering per level without extra passes.
Q2In a distributed system where tree nodes are stored across different services, how can you perform a level order traversal without pulling the entire tree into memory?
Leverage a producer‑consumer pattern where each service exposes an API to fetch a node’s immediate children; a central coordinator maintains a queue of node identifiers. It dequeues an identifier, requests its children, records the value, and enqueues the child identifiers. Because only one level’s identifiers are held at a time, memory usage stays proportional to the maximum width, matching the in‑memory BFS complexity.
Q3Why is a recursive solution for level order traversal generally discouraged in interview settings, and how can you still implement a recursive version efficiently?
Recursion naturally fits depth‑first traversals; a naïve recursive level order would require a separate pass for each depth, leading to O(n²) time. An efficient recursive version passes the current depth as a parameter and appends the node’s value to a list of lists indexed by depth, ensuring each node is visited once and achieving O(n) time, but interviewers often prefer the iterative queue version because it avoids stack overflow on deep trees and demonstrates explicit control of BFS.
Examples
Input
root = [1, 2, 3, null, 4, 5, 6]
Output
[[1], [2, 3], [4, 5, 6]]
Explanation: Level 0 contains the root node 1. Level 1 contains the left child 2 and right child 3. Level 2 contains the children of level 1: node 2 has a right child 4 (left is null), and node 3 has left child 5 and right child 6. Thus, level 2 is [4, 5, 6].
Input
root = [10, 20, 30, 40, null, 50, 60, null, 70]
Output
[[10], [20, 30], [40, 50, 60], [70]]
Explanation: Level 0: [10]. Level 1: [20, 30]. Level 2: Node 20 has left child 40; Node 30 has left child 50 and right child 60. So Level 2 is [40, 50, 60]. Level 3: Node 40 has right child 70; others are null. So Level 3 is [70].
Input
root = [5]
Output
[[5]]
Explanation: The tree consists of a single node. Level 0 contains only the root value 5. No further levels exist.
Input
root = null
Output
[]
Explanation: The input tree is empty. Therefore, there are no levels to traverse, and the result is an empty array.
Constraints
- The number of nodes in the tree is in the range [0, 10^4].
- -10^5 <= Node.val <= 10^5.
- The tree is a valid binary tree (each node has at most two children).
- The depth of the tree is at most 10^4.
Optimal Approach & Strategy
Use a single queue to perform BFS, processing nodes level by level in one pass. By capturing the queue size at the start of each level, you can directly build the per‑level lists without extra traversals, achieving O(n) time.
Brute Force Approach
A naive method would compute the height of the tree, then for each depth from 0 to height‑1 perform a separate DFS to collect nodes at that depth, resulting in repeated traversals. This leads to O(n²) time in the worst case because each node may be visited once per level.
Verified Code Solutions
function levelOrder(root) { let result = []; if (!root) return result; let queue = [root]; while (queue.length > 0) { let level = []; let levelSize = queue.length; for (let i = 0; i < levelSize; i++) { let node = queue.shift(); level.push(node.val); if (node.left) queue.push(node.left); if (node.right) queue.push(node.right); } result.push(level); } return result; }class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> result;
if (!root) return result;
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
vector<int> level;
int levelSize = q.size();
for (int i = 0; i < levelSize; i++) {
TreeNode* node = q.front();
q.pop();
level.push_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
result.push_back(level);
}
return result;
}
};import java.util.*;
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) return result;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
List<Integer> level = new ArrayList<>();
int levelSize = queue.size();
for (int i = 0; i < levelSize; i++) {
TreeNode node = queue.poll();
level.add(node.val);
if (node.left != null) queue.add(node.left);
if (node.right != null) queue.add(node.right);
}
result.add(level);
}
return result;
}
}from collections import deque
def levelOrder(root):
if not root:
return []
result = []
queue = deque([root])
while queue:
level = []
level_size = len(queue)
for _ in range(level_size):
node = queue.popleft()
level.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(level)
return resultfunction levelOrder(root) { let result = []; if (!root) return result; let queue = [root]; while (queue.length > 0) { let level = []; let levelSize = queue.length; for (let i = 0; i < levelSize; i++) { let node = queue.shift(); level.push(node.val); if (node.left) queue.push(node.left); if (node.right) queue.push(node.right); } result.push(level); } return result; }Asked in Top Tech Interviews
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