# 261 Graph Valid Tree

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.

For 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.

**The Idea:** A tree is a graph without cycle. So run a cyclic detection algorithm on the graph. Having an undirected graph simplifies this. Using a graph traversal all the needs to be checked is whether one path leads into an already visited node. This is because edges are bidirectional. Consider the case where this will not work with an undirected graph.

```
a----b
|     |
c----d

a---.b
|    .
.     |
c---.d
```

**Complexity:** O(n) time and space

```python
def validTree(self, n, edges):
        """
        :type n: int
        :type edges: List[List[int]]
        :rtype: bool
        """
        g = {i: set() for i in range(0, n)}
        for s, e in edges:
            g[s].add(e)
            g[e].add(s)

        def cyclic(root, n):
            visited = set()
            stack = [root]

            while stack:
                top = stack.pop()
                if top in visited:
                    return True
                visited.add(top)
                for neighbor in g[top]:
                    if (neighbor not in visited):
                        stack.append(neighbor)
            return len(visited) != n
        return not cyclic(0, n)
```


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