Spiral Traversal of 2D array in JavaScript

spiral-traversal.js

const test1 = [
    [1]
];
const test2 = [
    [1, 2],
    [3, 4]
];
const test3 = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9]
];
const test4 = [
    [1,2,3,4],
    [5,6,7,8],
    [9,1,2,3],
    [4,5,6,7]
];
const test5 = [
    [1,2,3,4,5],
    [6,7,8,9,0],
    [1,2,3,4,5],
    [6,7,8,9,0],
    [1,2,3,4,5]
];

// assume a square matrix
const spiralTraversal = (matrix, cb) => {
    let cycles = matrix.length;
    for (let cycle = 0; cycle < cycles; cycle++) {
        traverseCycle(matrix, cycle, cb);
    }
};

const traverseCycle = (matrix, cycle, cb) => {
    let indexedSize = matrix.length - 1;

    // we are at the midpoint of a odd sized matrix
    if (cycle + 0.5  === matrix.length/2) {
        cb(matrix[cycle][cycle]);
        return;
    }
    // going right
    for (let row = cycle, col = 0 + cycle; col < indexedSize - cycle; col++) {
        cb(matrix[row][col]);
    }
    // going down
    for (let row = cycle, col = indexedSize - cycle; row < indexedSize - cycle; row++) {
        cb(matrix[row][col]);
    }
    // going left
    for (let row = indexedSize - cycle, col = indexedSize - cycle; col > cycle; col--) {
        cb(matrix[row][col]);
    }
    // going up
    for (let row = indexedSize - cycle, col = cycle; row > cycle; row--) {
        cb(matrix[row][col]);
    }
};