LintCode_178 Graph Valid Tree

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

 Notice

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.

Example

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

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

思路是用并查集实现，如果一条边的两个点有相同的parent，意味着两个点属于一个集合， 所以必回形成环，所以就不是tree；

最后还要检查下有几个集合， 如果多余一个， 说明没连起来， 索爷也不是tree；

class Solution {public:    /**     * @param n an integer     * @param edges a list of undirected edges     * @return true if it's a valid tree, or false     */    bool validTree(int n, vector<vector<int>>& edges) {        // Write your code here        vector<int> linkedset;        if (n == 1 && edges.empty())            return true;        for (int i = 0; i < n; i++) {            linkedset.push_back(i);        }        for (int i = 0; i < edges.size(); i++) {            int root1= get_root(edges[i][0], linkedset);            int root2= get_root(edges[i][1], linkedset);            if (root1 == root2)                return false;            if (root1 > root2) {                linkedset[root1] = root2;            } else {                linkedset[root2] = root1;            }        }        int num_of_root = 0;        for (int i = 0; i < linkedset.size(); i++) {            if (linkedset[i] == i)               num_of_root++;        }        if (num_of_root != 1)            return false;        else            return true;    }        int get_root(int i, vector<int>& linkedset) {        while (linkedset[i] != i) {            i = linkedset[i];        }        return i;    }};