Leetcode - 102. Binary Tree Level Order Traversal

🧠 Approach: Breadth-First Search (BFS)

This is a textbook case for BFS using a queue. You process nodes level by level, tracking all nodes at each depth before moving to the next.

Steps:

1. Use a queue initialized with the root node.
2. At each level:

• Record all node values into an array.
• Add their children to the queue.
  1. Push the level array into the final result.

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✅ Code (JavaScript):

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val === undefined ? 0 : val);
 *     this.left = (left === undefined ? null : left);
 *     this.right = (right === undefined ? null : right);
 * }
 */

/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var levelOrder = function(root) {
  if (root === null) return [];

  const result = [];
  const queue = [root];

  while (queue.length > 0) {
    const levelSize = queue.length;
    const level = [];

    for (let i = 0; i < levelSize; i++) {
      const 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;
};

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🧮 Example:

Input Tree:

       3
      / \
     9  20
        / \
       15  7

Output:

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

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⏱️ Time & Space Complexity

Type	Complexity
Time	O(n) — Visit every node once
Space	O(n) — Queue and result array can hold all nodes

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