102. Binary Tree Level Order Traversal

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).

Example 1:

Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]]
Example 2:
Input: root = [1] Output: [[1]]
Example 3:
Input: root = [] Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

Runtime: 80 ms, faster than 82.71% of JavaScript online submissions for Binary Tree Level Order Traversal.
Memory Usage: 40.9 MB, less than 7.95% of JavaScript online submissions for Binary Tree Level Order Traversal.

/**
 * 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) {
        return [];
    }
    const result = [];
    const queue = [root];
    let level;
    let node;
    let n;
    while(queue.length > 0) {
        level = [];
        n = queue.length;
        for (let i = 0; i < n; i++) {
            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;
};