# Solution For Graph Valid Tree

The Graph Valid Tree problem on LeetCode asks whether a given graph is a valid tree. A graph is a valid tree if it is undirected, connected, and has no cycles.

To solve this problem, we can use Depth First Search (DFS) or Breadth First Search (BFS) to traverse the graph and check if it meets the conditions of a valid tree.

Algorithm:

- Create a visited set to keep track of nodes we have already visited.
- Create an adjacency list to represent the graph.
- Initialize a queue or stack (depending on whether we use BFS or DFS) with the starting node.
- While the queue or stack is not empty, do the following:

a. Pop the next node from the queue or stack.

b. If the node is already in the visited set, return False, because we have found a cycle.

c. Otherwise, add the node to the visited set and add its neighbors to the queue or stack. - After traversing the entire graph, if all nodes have been visited and there are no cycles, return True. Otherwise, return False.

Code Implementation:

Using DFS:

`class Solution:`

def validTree(self, n: int, edges: List[List[int]]) -> bool:

# create visited set and adjacency list

visited = set()

adj_list = {i: [] for i in range(n)}

for u, v in edges:

adj_list[u].append(v)

adj_list[v].append(u)

def dfs(node, parent):

# if the node is already visited, return False

if node in visited:

return False

# add node to visited set and explore its neighbors

visited.add(node)

for neighbor in adj_list[node]:

# skip parent of current node to avoid cycle

if neighbor == parent:

continue

# if neighbor has cycle, return False

if not dfs(neighbor, node):

return False

return True

# start the DFS traversal from the first node

return dfs(0, -1) and len(visited) == n

The time complexity of this DFS solution is O(V+E) where V is the number of vertices and E is the number of edges.

Using BFS:

`class Solution:`

def validTree(self, n: int, edges: List[List[int]]) -> bool:

# create visited set, adjacency list, and initial queue

visited = set()

adj_list = {i: [] for i in range(n)}

queue = collections.deque([0])

for u, v in edges:

adj_list[u].append(v)

adj_list[v].append(u)

# BFS traversal

while queue:

node = queue.popleft()

if node in visited:

return False

visited.add(node)

for neighbor in adj_list[node]:

queue.append(neighbor)

# remove the edge between node and neighbor

adj_list[neighbor].remove(node)

return len(visited) == n

The time complexity of this BFS solution is also O(V+E) where V is the number of vertices and E is the number of edges.

## Step by Step Implementation For Graph Valid Tree

There are a few different ways to approach this problem. One way would be to use a depth-first search algorithm to traverse the graph and keep track of which nodes have been visited. If a node is visited twice, then there is a cycle and the graph is not valid. Another way would be to use a breadth-first search algorithm and keep track of the distance from the starting node to each node that is visited. If two nodes are the same distance away from the starting node, then there is a cycle and the graph is not valid.

def validTree(n, edges): # initialize n isolated islands nums = [i for i in range(n)] # perform union find for e in edges: root1 = find(nums, e[0]) root2 = find(nums, e[1]) # if two vertices happen to be in the same set # then there's a cycle if root1 == root2: return False # union nums[root2] = root1 return len(edges) == n-1 def find(nums, i): if nums[i] == i: return i return find(nums, nums[i])

There are a few different ways to approach this problem. One way would be to use a depth-first search algorithm. Another way would be to use a breadth-first search algorithm. We can also use a Union-Find algorithm. Here is a sample solution using a Union-Find algorithm: var find = function(x) { if(parents[x] == x) { return x; } return find(parents[x]); } var union = function(x, y) { var rootx = find(x); var rooty = find(y); if(rootx != rooty) { parents[rooty] = rootx; } } var isValidTree = function(n, edges) { var parents = []; for(var i = 0; i < n; i++) { parents[i] = i; } for(var i = 0; i < edges.length; i++) { var x = edges[i][0]; var y = edges[i][1]; if(find(x) == find(y)) { return false; } union(x, y); } return edges.length == n - 1; };

There are a few different ways to approach this problem. One way would be to use a depth-first search algorithm to determine if there is a path between all of the nodes in the graph. If there is a path between all of the nodes, then the graph is considered valid. Another way to approach this problem would be to use a breadth-first search algorithm. This algorithm would also determine if there is a path between all of the nodes in the graph. If there is a path, then the graph is considered valid.

using System; using System.Collections.Generic; public class Solution { public bool ValidTree(int n, int[,] edges) { // check for extreme conditions if (n == 0) { return true; } if (edges.GetLength(0) != n - 1) { return false; } // build the adjacency list List[] adjList = new List [n]; for (int i = 0; i < n; i++) { adjList[i] = new List (); } for (int i = 0; i < n - 1; i++) { int u = edges[i, 0]; int v = edges[i, 1]; adjList[u].Add(v); adjList[v].Add(u); } // use a visited array to track nodes that have been visited bool[] visited = new bool[n]; // do a depth first search from node 0 if (!DoDFS(0, -1, ref visited, adjList)) { return false; } // check that all nodes were visited for (int i = 0; i < n; i++) { if (!visited[i]) { return false; } } return true; } private bool DoDFS(int curr, int parent, ref bool[] visited, List [] adjList) { // mark the current node as visited visited[curr] = true; // visit each neighbor foreach (int neighbor in adjList[curr]) { // if the neighbor has been visited and it is not the parent, then there is a cycle if (visited[neighbor] && neighbor != parent) { return false; } // if the neighbor has not been visited, do a depth first search on it if (!DoDFS(neighbor, curr, ref visited, adjList)) { return false; } } return true; } }