[LeetCode] Graph Valid Tree [Union Find]

104828720 发布于2019-08-14 15:10 / 2673人阅读

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.

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.

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.

Solution

public class Solution {
    
    int[] parents;
    public boolean validTree(int n, int[][] edges) {
        if (n-1 != edges.length) {
            return false;
        }
        parents = new int[n];
        //initialize n nodes as each own parent
        for (int i = 0; i < n; i++) {
            parents[i] = i;
        }
        for (int i = 0; i < n-1; i++) {
            if (find(edges[i][0]) == find(edges[i][1])) {
                return false; //if two nodes on edge[i] have the same parent, this edge makes a circle causing it invalid
            }
            else union(edges[i][0], edges[i][1]); //else union the two nodes on edge[i]
        }
        return true; 
    }
    public int find(int node) {
        //find the very first parent node, which is when the parent is the node itself
        if (parents[node] == node) {
            return node;
        }
        //find parent recursively
        return find(parents[node]);
    }
    public void union(int a, int b) {
        int finda = parents[a], findb = parent[b];
        //when node a and node b have different ancient parents, union them with the same parent
        if (finda != findb) {
            parents[finda] = findb;
        }
    }
}

Update 2018-10

class Solution {
    public boolean validTree(int n, int[][] edges) {
        if (edges.length != n-1) return false;
        int[] nums = new int[n];
        Arrays.fill(nums, -1);
        for (int i = 0; i < edges.length; i++) {
            int x = find(nums, edges[i][0]);
            int y = find(nums, edges[i][1]);
            if (x == y) return false;
            nums[x] = y;
        }
        return true;
    }
    private int find(int[] nums, int k) {
        if (nums[k] == -1) return k;
        else return find(nums, nums[k]);
    }
}

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