[LintCode]图是否是树

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http://www.lintcode.com/zh-cn/problem/graph-valid-tree/#

解法一：并查集

 id[i]保存当前结点i的根节点，因为如果有公共根节点，那么再相连就一定成环，每次find的时候将不是root的置为root

public class Solution {    /**     * @param n an integer     * @param edges a list of undirected edges     * @return true if it's a valid tree, or false     */    // id保存当前结点的根节点，因为如果有公共根节点，那么再相连就一定成环    int[] id;    int count;    public boolean validTree(int n, int[][] edges) {        // Write your code here        count = n;        id = new int[n];        for (int i = 0; i < n; i++) {            id[i] = i;        }        for (int i = 0; i < edges.length; i++) {            int p = edges[i][0];            int q = edges[i][1];            if (find(p) == find(q)) {                return false;            } else {                union(p, q);            }        }        return count == 1;    }    private int find(int i) {        int root = i;        while (root != id[root]) {            root = id[root];        }        while (i != root) {            int tmp = id[i];            id[i] = root;            i = tmp;        }        return root;    }    private void union(int i, int j) {        int p = find(i);        int q = find(j);        if (p != q) {            id[p] = q;            count--;        }    }}

解法二：DFS用List<List<Integer>>保存连接关系，visited保存已访问位置，如果将要访问的在已访问的之中，那么就有环。

public class Solution {    /**     * @param n an integer     * @param edges a list of undirected edges     * @return true if it's a valid tree, or false     */    public boolean validTree(int n, int[][] edges) {        // Write your code here        HashSet<Integer> visited = new HashSet<>();        List<List<Integer>> list = new LinkedList<>();        for (int i = 0; i < n; i++) {            list.add(new ArrayList<Integer>());        }        for (int i = 0; i < edges.length; i++) {            list.get(edges[i][0]).add(edges[i][1]);            list.get(edges[i][1]).add(edges[i][0]);        }        if (cycle(-1, 0, list, visited)) {            return false;        }        return visited.size() == n;    }    private boolean cycle(Integer pre, Integer cur, List<List<Integer>> list, HashSet<Integer> visit) {        visit.add(cur);        for (Integer succ : list.get(cur)) {            if (!succ.equals(pre)) {                if (visit.contains(succ)) {                    return true;                } else {                    if (cycle(cur, succ, list, visit)) {                        return true;                    }                }            }        }        return false;    }}

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