【问题描述】[中等]

【解答思路】

1. DFS 深度优先遍历

时间复杂度：O(N+M) 空间复杂度：O(N)

class Solution {private static final int UNCOLORED = 0;private static final int RED = 1;private static final int GREEN = 2;private int[] color;private boolean valid;public boolean isBipartite(int[][] graph) {int n = graph.length;valid = true;color = new int[n];Arrays.fill(color, UNCOLORED);for (int i = 0; i < n && valid; ++i) {if (color[i] == UNCOLORED) {dfs(i, RED, graph);}}return valid;}public void dfs(int node, int c, int[][] graph) {color[node] = c;int cNei = c == RED ? GREEN : RED;for (int neighbor : graph[node]) {if (color[neighbor] == UNCOLORED) {dfs(neighbor, cNei, graph);if (!valid) {return;}} else if (color[neighbor] != cNei) {valid = false;return;}}}
}

2. 广度优先遍历 BFS

时间复杂度：O(N+M) 空间复杂度：O(N)

class Solution {private static final int UNCOLORED = 0;private static final int RED = 1;private static final int GREEN = 2;private int[] color;public boolean isBipartite(int[][] graph) {int n = graph.length;color = new int[n];Arrays.fill(color, UNCOLORED);for (int i = 0; i < n; ++i) {if (color[i] == UNCOLORED) {Queue<Integer> queue = new LinkedList<Integer>();queue.offer(i);color[i] = RED;while (!queue.isEmpty()) {int node = queue.poll();int cNei = color[node] == RED ? GREEN : RED;for (int neighbor : graph[node]) {if (color[neighbor] == UNCOLORED) {queue.offer(neighbor);color[neighbor] = cNei;} else if (color[neighbor] != cNei) {return false;}}}}}return true;}
}

【总结】

1. 深度优先遍历 栈/递归 广度优先遍历 队列

2有向图、无向图 、树 ----> BFS DFS思想

转载链接：https://leetcode-cn.com/problems/is-graph-bipartite/solution/pan-duan-er-fen-tu-by-leetcode-solution/