DSA interview question

Binary Tree Level Order Traversal

Easy · Data structures · Topics: tree, bfs, queue

The problem

An org-chart viewer displays a company tree level by level: the CEO, then all direct reports, then their reports, and so on, each level as its own row. For a binary tree, that's a breadth-first search with a queue, where each iteration of the outer loop processes exactly the nodes currently in the queue, which are exactly one level.

Example: [3, 9, 20, null, null, 15, 7] → [[3], [9, 20], [15, 7]].

Function signature

vector<vector<int>> levelOrder(TreeNode* root);

Test cases

Input (level order)	Output	Explanation
`[3, 9, 20, null, null, 15, 7]`	`[[3], [9, 20], [15, 7]]`
`[1]`	`[[1]]`
`[]`	`[]`
`[1, 2, 3, 4, null, null, 5]`	`[[1], [2, 3], [4, 5]]`	Gaps don't matter
`[1, 2, null, 3]`	`[[1], [2], [3]]`	A left chain

(Confirm: each inner list lists one level from left to right.)

How interviewers phrase it

"Given the root of a binary tree, return the level order traversal of its nodes' values, level by level from left to right" (LeetCode 102).

Clarifying questions to ask

One list per level, or a flat list? (One per level.)
Left to right within a level? (Yes; zigzag is a separate problem.)
Can the tree be empty? (Yes, so return an empty list.)

Follow-ups to expect

"How do you separate levels?"
"DFS alternative?"
"Related variations?"
"Space?"
"Complexity?"

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