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[LeetCode] Graph Valid Tree 图验证树

 

Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.

For example:

Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return true.

Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return false.

Hint:

  1. Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], what should your return? Is this case a valid tree?
  2. According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”

Note: you can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

 

这道题给了我们一个无向图,让我们来判断其是否为一棵树,我们知道如果是树的话,所有的节点必须是连接的,也就是说必须是连通图,而且不能有环,所以我们的焦点就变成了验证是否是连通图和是否含有环。我们首先用DFS来做,根据pair来建立一个图的结构,用邻接链表来表示,还需要一个一位数组v来记录某个节点是否被访问过,然后我们用DFS来搜索节点0,遍历的思想是,当DFS到某个节点,先看当前节点是否被访问过,如果已经被访问过,说明环存在,直接返回false,如果未被访问过,我们现在将其状态标记为已访问过,然后我们到邻接链表里去找跟其相邻的节点继续递归遍历,注意我们还需要一个变量pre来记录上一个节点,以免回到上一个节点,这样遍历结束后,我们就把和节点0相邻的节点都标记为true,然后我们在看v里面是否还有没被访问过的节点,如果有,则说明图不是完全连通的,返回false,反之返回true,参见代码如下:

 

解法一:

// DFS
class Solution {
public:
    bool validTree(int n, vector<pair<int, int>>& edges) {
        vector<vector<int>> g(n, vector<int>());
        vector<bool> v(n, false);
        for (auto a : edges) {
            g[a.first].push_back(a.second);
            g[a.second].push_back(a.first);
        }
        if (!dfs(g, v, 0, -1)) return false;
        for (auto a : v) {
            if (!a) return false;
        }
        return true;
    }
    bool dfs(vector<vector<int>> &g, vector<bool> &v, int cur, int pre) {
        if (v[cur]) return false;
        v[cur] = true;
        for (auto a : g[cur]) {
            if (a != pre) {
                if (!dfs(g, v, a, cur)) return false;
            }
        }
        return true;
    }
};

 

下面我们来看BFS的解法,思路很相近,需要用queue来辅助遍历,这里我们没有用一维向量来标记节点是否访问过,而是用了一个set,如果遍历到一个节点,在set中没有,则加入set,如果已经存在,则返回false,还有就是在遍历邻接链表的时候,遍历完成后需要将节点删掉,参见代码如下:

 

解法二:

// BFS
class Solution {
public:
    bool validTree(int n, vector<pair<int, int>>& edges) {
        vector<unordered_set<int>> g(n, unordered_set<int>());
        unordered_set<int> s{{0}};
        queue<int> q{{0}};
        for (auto a : edges) {
            g[a.first].insert(a.second);
            g[a.second].insert(a.first);
        }
        while (!q.empty()) {
            int t = q.front(); q.pop();
            for (auto a : g[t]) {
                if (s.count(a)) return false;
                s.insert(a);
                q.push(a);
                g[a].erase(t);
            }
        }
        return s.size() == n;
    }
};

 

我们再来看Union Find的方法,这种方法对于解决连通图的问题很有效,思想是我们遍历节点,如果两个节点相连,我们将其roots值连上,这样可以帮助我们找到环,我们初始化roots数组为-1,然后对于一个pair的两个节点分别调用find函数,得到的值如果相同的话,则说明环存在,返回false,不同的话,我们将其roots值union上,参见代码如下:

 

解法三:

// Union Find
class Solution {
public:
    bool validTree(int n, vector<pair<int, int>>& edges) {
        vector<int> roots(n, -1);
        for (auto a : edges) {
            int x = find(roots, a.first), y = find(roots, a.second);
            if (x == y) return false;
            roots[x] = y;
        }
        return edges.size() == n - 1;
    }
    int find(vector<int> &roots, int i) {
        while (roots[i] != -1) i = roots[i];
        return i;
    }
};

 

类似题目:

Number of Islands II

Number of Connected Components in an Undirected Graph

 

参考资料:

https://leetcode.com/discuss/85398/bfs-java-solution

https://leetcode.com/discuss/80142/my-c-union-find-code

https://leetcode.com/discuss/52563/ac-java-union-find-solution

https://leetcode.com/discuss/86035/c-dfs-with-adjacent-list-graph

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-03-09 13:56 Grandyang 阅读(...) 评论(...) 编辑 收藏