261. Graph Valid Tree

最新推荐文章于 2024-04-18 23:32:51 发布

原创最新推荐文章于 2024-04-18 23:32:51 发布 · 710 阅读

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#union-find

本文介绍了一种使用Union-Find算法来判断一组无向边是否构成有效树的方法。通过实现Quick-Union算法，确保了图中不存在环且所有节点都是连通的。文中提供了一个具体的解决方案，并通过示例展示了算法的工作原理。

Problem

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.

Show Hint

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 inedges.

Solution

想到union find 就好办了。

实现 union - find , 有两种实现方式，quick-union 和 quick-find ( 参看这里 )

quick-union 实现比较简单 : 初始化大小为n, default value = -1的数组 arr, arr[ i ]= j 就表示 i 的父节点是 j ，

这样如果两个数字的父节点相同就说明在同一个union里；参考下图

Solution

class Solution {
    
    int find(const vector<int>& arr, int e){
        if(arr[e] == -1){
            return e;
        }
        return find(arr, arr[e]);
    }
public:
    bool validTree(int n, vector<pair<int, int>>& edges) {
        
        vector<int> root(n, -1);
        for( auto edge : edges){
            int set1 = find(root, edge.first);
            int set2 = find(root, edge.second);
            if(set1 == set2) 
                 return false;
            root[set1] = set2;
        }
        return edges.size() == n -1; // <span style="color:#ff0000;">buggy 不能简单的return true. e.g. edge太少，没能连通。</span>
    }
};