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🌳 Tree: Binary Tree Level Order Traversal

📝 Description

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

Real-World Application

This is Breadth-First Search (BFS). It's used in Social Networks (finding friends of friends), Web Crawlers (visiting links layer by layer), and broadcasting in networks.

🛠️ Constraints & Edge Cases

  • Number of nodes is between 0 and \(10^4\).
  • Edge Cases to Watch:
    • Empty tree (return []).
    • Skewed tree (still processes level by level).

🧠 Approach & Intuition

The Aha! Moment

We need to process nodes row by row. A Queue (FIFO) is perfect for this. We put the root in. Then, for every node we take out, we put its children in the back. To separate levels, we can snapshot the queue size at the start of each level loop.

🐢 Brute Force (Naive)

DFS with a depth parameter, adding to res[depth]. This works (\(O(N)\)) but isn't "level order" traversal logic, it's just organizing DFS output.

🐇 Optimal Approach

  1. Initialize q with root.
  2. While q is not empty:
    • Get len(q) (this is the number of nodes in the current level).
    • Iterate that many times:
      • Pop node.
      • Add value to level_list.
      • Add children to q.
    • Append level_list to result.

🧩 Visual Tracing

graph TD
    Q[Queue: 3] -->|Pop 3| L1[Level 1: 3]
    L1 -->|Push 9, 20| Q2[Queue: 9, 20]
    Q2 -->|Pop 9, 20| L2[Level 2: 9, 20]
    L2 -->|Push 15, 7| Q3[Queue: 15, 7]

💻 Solution Implementation

(Implementation details need to be added...)

⏱️ Complexity Analysis

  • Time Complexity: \(\mathcal{O}(N)\) — Visit every node once.
  • Space Complexity: \(\mathcal{O}(N)\) — The queue can hold up to \(N/2\) nodes (leaf level).

🎤 Interview Toolkit

  • Harder Variant: ZigZag Level Order Traversal.
  • Alternative: Recursive DFS passing level index (Pre-order).