If there's only two teams and they went 1-1 against each other they could play a direct tiebreaker, IDK what happens if there are three 4-2s and an 0-6 though. (Either they would have all gone 1-1 against each other, or each would have beaten one of the others 2-0 in a triangle, either way, direct matchups would be useless as a tiebreaker.)
Edited to add: the official rules are available on the web, turns out in a 3-way tie the teams would be seeded into a mini-gauntlet based on the total game times of each team's wins. The two teams (of the tied ones) that took the longest to win would play each other and then the winner would play the team with the shortest wins. If it's three 4-2s and a 0-6 the one with the shortest win times would thus clinch advancement and only have to play for seeding, but if it's one 6-0 and three 2-4s, then only one of the three tied teams, the overall winner of the mini-gauntlet, would advance at all.
If all four teams are 3-3, then all four get seeded into a single elimination bracket based on win times, the winners of the first two tiebreakers play each other for seeding into the main (Bo5) bracket and the losers are eliminated.
All tiebreakers are played as single games, with side selection going to the team with the fastest wins.
The rules don't specifically address the question of what if two teams have exactly the same total win length (becomes a bit less unlikely if game lengths are only recorded to the nearest second, although in theory Riot could probably record in a smaller time increment).