Question

Formatted question description: https://leetcode.ca/all/261.html

You have a graph of n nodes labeled from 0 to n - 1. You are given an integer n and a list of edges where edges[i] = [a_i, b_i] indicates that there is an undirected edge between nodes a_i and b_i in the graph.

Return true if the edges of the given graph make up a valid tree, and false otherwise.

Example 1:

Input: n = 5, edges = [[0,1],[0,2],[0,3],[1,4]]
Output: true

Example 2:

Input: n = 5, edges = [[0,1],[1,2],[2,3],[1,3],[1,4]]
Output: false

Constraints:

1 <= n <= 2000
0 <= edges.length <= 5000
edges[i].length == 2
0 <= a_i, b_i < n
a_i != b_i
There are no self-loops or repeated edges.

Algorithm

If it is a tree:

all nodes must be connected, that is, it must be a connected graph
and there must be no cycles

Use queue to assist the traversal. Here, instead of using a one-dimensional vector to mark whether the node has been visited, a HashSet is used. If a node is traversed and there is no HashSet in the HashSet, the HashSet is added. If it already exists, false is returned.

Code

import java.util.*;

public class Graph_Valid_Tree {

    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) {
            if (n == 0) {
                return false;
            }

            if (edges.length != n - 1) { // quick check
                return false;
            }

            Map<Integer, Set<Integer>> graph = initializeGraph(n, edges);

            // bfs
            Queue<Integer> queue = new LinkedList<>();
            Set<Integer> isVisited = new HashSet<>();

            queue.offer(0);
            isVisited.add(0);
            while (!queue.isEmpty()) {
                int node = queue.poll();
                for (Integer neighbor : graph.get(node)) {
                    if (isVisited.contains(neighbor)) { // cycle found
                        return false;
                    }
                    isVisited.add(neighbor);
                    queue.offer(neighbor);
                }
            }

            return (isVisited.size() == n); // if all nodes connected
        }

        private Map<Integer, Set<Integer>> initializeGraph(int n, int[][] edges) {

            // node => its neibougher
            Map<Integer, Set<Integer>> graph = new HashMap<>();
            for (int i = 0; i < n; i++) {
                graph.put(i, new HashSet<Integer>());
            }

            for (int i = 0; i < edges.length; i++) {
                int u = edges[i][0];
                int v = edges[i][1];
                graph.get(u).add(v);
                graph.get(v).add(u);
            }

            return graph;
        }
    }
}

############

class Solution {
    private int[] p;

    public boolean validTree(int n, int[][] edges) {
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            if (find(a) == find(b)) {
                return false;
            }
            p[find(a)] = find(b);
            --n;
        }
        return n == 1;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

// OJ: https://leetcode.com/problems/graph-valid-tree/
// Time: O(N + E)
// Space: O(N)
class UnionFind {
    vector<int> id;
    int cnt;
public:
    UnionFind(int N) : id(N), cnt(N) {
        iota(begin(id), end(id), 0);
    }
    int find(int x) {
        return id[x] == x ? x : (id[x] = find(id[x]));
    }
    void connect(int a, int b) {
        id[find(a)] = find(b);
        --cnt;
    }
    bool connected(int a, int b) {
        return find(a) == find(b);
    }
    int getCount() { return cnt; }
};
class Solution {
public:
    bool validTree(int n, vector<vector<int>>& E) {
        UnionFind uf(n);
        for (auto &e : E) {
            if (uf.connected(e[0], e[1])) return false;
            uf.connect(e[0], e[1]);
        }
        return uf.getCount() == 1;
    }
};

class Solution:
    def validTree(self, n: int, edges: List[List[int]]) -> bool:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(n))
        for a, b in edges:
            if find(a) == find(b):
                return False
            p[find(a)] = find(b)
            n -= 1
        return n == 1

############

class Solution:
  # @param {int} n an integer
  # @param {int[][]} edges a list of undirected edges
  # @return {boolean} true if it's a valid tree, or false
  def validTree(self, n, edges):
    # Write your code here

    def dfs(root, graph, visited, parent):
      visited[root] = 1
      for nbr in graph.get(root, []):
        if nbr == parent:
          continue
        elif visited[nbr] != 0:
          return False
        if not dfs(nbr, graph, visited, root):
          return False
      visited[root] = 2
      self.nodeVisited += 1
      return True

    visited = [0 for _ in range(n)]
    graph = {}
    self.nodeVisited = 0
    for edge in edges:
      start, end = edge[0], edge[1]
      graph[start] = graph.get(start, []) + [end]
      graph[end] = graph.get(end, []) + [start]

    if dfs(0, graph, visited, -1) and self.nodeVisited == n:
      return True
    else:
      return False

func validTree(n int, edges [][]int) bool {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for _, e := range edges {
		a, b := e[0], e[1]
		if find(a) == find(b) {
			return false
		}
		p[find(a)] = find(b)
		n--
	}
	return n == 1
}

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {boolean}
 */
var validTree = function (n, edges) {
    let p = new Array(n);
    for (let i = 0; i < n; ++i) {
        p[i] = i;
    }
    function find(x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
    for (const [a, b] of edges) {
        if (find(a) == find(b)) {
            return false;
        }
        p[find(a)] = find(b);
        --n;
    }
    return n == 1;
};

261 - Graph Valid Tree

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