Javascript Tree Traversal Algorithm

Question

Necesito ayuda para atravesar una estructura de árbol de una manera profunda. No puedo idear un algoritmo para hacerlo correctamente.

Mi entrada es la siguiente:

[ 
    ["A", "B", "C"], 
    ["1", "2"], 
    ["a", "b", "c", "d"] 
] 

La salida debe tomar la forma:

[ 
    "A/1/a", "A/1/b", "A/1/c", "A/1/d", 
    "A/2/a", "A/2/b", "A/2/c", "A/2/d", 
    "B/1/a", "B/1/b", "B/1/c", "B/1/d", 
    "B/2/a", "B/2/b", "B/2/c", "B/2/d", 
    "C/1/a", "C/1/b", "C/1/c", "C/1/d", 
    "C/2/a", "C/2/b", "C/2/c", "C/2/d" 
] 

Answer 1

Esto debería hacer el trabajo:

function traverse(arr) { 
 
    var first = arr[0]; 
 
    var ret = []; 
 
    if (arr.length == 1) { 
 
    for (var i = 0; i < first.length; i++) { 
 
     ret.push(first[i]); 
 
    } 
 
    } else { 
 
    for (var i = 0; i < first.length; i++) { 
 
     var inn = traverse(arr.slice(1)); 
 
     for (var j = 0; j < inn.length; j++) { 
 
     ret.push(first[i] + '/' + inn[j]); 
 
     } 
 
    } 
 
    } 
 
    return ret; 
 
} 
 

 
var inp = [ 
 
    ["A", "B", "C"], 
 
    ["1", "2"], 
 
    ["a", "b", "c", "d"] 
 
]; 
 

 
var out = traverse(inp); 
 

 
console.log(out);

Answer 2

Lo que estamos buscando es el producto cartesiano de una lista de listas, que ha sido asked antes de. Los préstamos de la respuesta aceptada por esa pregunta, usted puede hacer esto en Javascript 1.7:

function product() { 
    return Array.prototype.reduce.call(arguments, function(as, bs) { 
     return [a.concat(b) for each (a in as) for each (b in bs)]; 
    }, [[]]); 
}; 

function convert(lst) { 
    var solution = []; 
    for (var i = 0; i < lst.length; i++) { 
    solution.push(lst[i][0] + "/" + lst[i][1] + "/" + lst[i][2]); 
    } 
    return solution; 
}; 

convert(product(["A", "B", "C"], ["1", "2"], ["a", "b", "c", "d"])); 

> ["A/1/a", "A/1/b", "A/1/c", "A/1/d", 
    "A/2/a", "A/2/b", "A/2/c", "A/2/d", 
    "B/1/a", "B/1/b", "B/1/c", "B/1/d", 
    "B/2/a", "B/2/b", "B/2/c", "B/2/d", 
    "C/1/a", "C/1/b", "C/1/c", "C/1/d", 
    "C/2/a", "C/2/b", "C/2/c", "C/2/d"]