A bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint and independent sets U and V such that every edge connects a vertex in U to one in V. Vertex sets U and V are usually called the parts of the graph. Equivalently, a bipartite graph is a graph that does not contain any odd-length cycles.
Bipartite
Is graph bipartite?
DFS
DFS to jump between two parts and assign part number to nodes
Take each node in the Graph as starting point of DFS.
Assign the nodes to one of two parts, -1 and 1, in an alternating order.
If a node is already assigned to a part, once it's visited again, its part should be the same as the part of the previous node.
// OJ: https://leetcode.com/problems/is-graph-bipartite/
// Author: github.com/lzl124631x
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
bool isBipartite(vector<vector<int>>& G) {
vector<int> id(G.size());
function<bool(int, int)> dfs = [&](int u, int prev) {
if (id[u]) return id[u] != prev; // This node's part should be the same as the part of the previous part
id[u] = -prev; // Assign nodes to one of the two parts, -1 or 1
for (int v : G[u]) {
if (!dfs(v, id[u])) return false;
}
return true;
};
for (int i = 0; i < G.size(); ++i) {
if (id[i]) continue; // This node is already assigned to a part
if (!dfs(i, 1)) return false;
}
return true;
}
};
BFS
Take each node in the Graph as starting point of BFS.
Assign nodes to part -1 or 1 layer by layer in an alternating order.
If a node is already assigned to a part, once it's visited again, its part should be the same as the part of the previous node.
// OJ: https://leetcode.com/problems/is-graph-bipartite/
// Author: github.com/lzl124631x
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
bool isBipartite(vector<vector<int>>& G) {
vector<int> id(G.size());
queue<int> q;
for (int i = 0; i < G.size(); ++i) {
if (id[i]) continue; // This node is already assigned to a part
q.push(i);
id[i] = 1; // Assign this starting node to part 1
while (q.size()) {
int u = q.front();
q.pop();
for (int v : G[u]) {
if (id[v]) {
if (id[v] != -id[u]) return false; // Two neighboring nodes shouldn't be in the same part
continue;
}
id[v] = -id[u]; // Assign neighbor node to the other part
q.push(v);
}
}
}
return true;
}
};
