261. Graph Valid Tree

Explanation:

To determine if a given graph is a valid tree, we need to check two conditions:

The graph is fully connected (no isolated nodes).
The graph has no cycles.

We can solve this problem using a depth-first search (DFS) algorithm. We will start the DFS traversal from any node and visit all its neighbors. While visiting each neighbor, we will mark the current node as visited. If we encounter a visited node, it means there is a cycle in the graph. Additionally, we will keep track of the number of connected components in the graph. If there is more than one connected component, the graph is not a valid tree.

: :

View Leetcode Problem

class Solution {
    public boolean validTree(int n, int[][] edges) {
        List<List<Integer>> adjList = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adjList.add(new ArrayList<>());
        }

        for (int[] edge : edges) {
            adjList.get(edge[0]).add(edge[1]);
            adjList.get(edge[1]).add(edge[0]);
        }

        boolean[] visited = new boolean[n];
        if (hasCycle(-1, 0, visited, adjList)) {
            return false;
        }

        for (boolean v : visited) {
            if (!v) {
                return false;
            }
        }

        return true;
    }

    private boolean hasCycle(int parent, int node, boolean[] visited, List<List<Integer>> adjList) {
        visited[node] = true;

        for (int neighbor : adjList.get(node)) {
            if (!visited[neighbor]) {
                if (hasCycle(node, neighbor, visited, adjList)) {
                    return true;
                }
            } else if (neighbor != parent) {
                return true;
            }
        }

        return false;
    }
}

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