Is graph bipartite

/*
 * There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties:
 * There are no self-edges (graph[u] does not contain u).
 * There are no parallel edges (graph[u] does not contain duplicate values).
 * If v is in graph[u], then u is in graph[v] (the graph is undirected).
 * The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them.
 * A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B.
 * Return true if and only if it is bipartite.
 * Example 1:
 * Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
 * Output: false
 * Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.
 * Example 2:
 * Input: graph = [[1,3],[0,2],[1,3],[0,2]]
 * Output: true
 * Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.
 * Constraints:
 * graph.length == n
 * 1 <= n <= 100
 * 0 <= graph[u].length < n
 * 0 <= graph[u][i] <= n - 1
 * graph[u] does not contain u.
 * All the values of graph[u] are unique.
 * If graph[u] contains v, then graph[v] contains u.
 */
impl Solution {
  pub fn is_bipartite(graph: Vec<Vec<i32>>) -> bool {
    let n = graph.len();
    let mut color = vec![-1i32; n];
    for start in 0..n {
      if color[start] != -1 { continue; }
      let mut queue = std::collections::VecDeque::new();
      queue.push_back(start);
      color[start] = 0;
      while let Some(u) = queue.pop_front() {
        for &v in &graph[u] {
          let v = v as usize;
          if color[v] == -1 {
            color[v] = 1 - color[u];
            queue.push_back(v);
          } else if color[v] == color[u] {
            return false;
          }
        }
      }
    }
    true
  }
}