Graph Valid Tree
A graph is a valid tree if and only if it has exactly ( n  1 ) edges, and there is a way to traverse from one node to any other node. If a graph has more than or fewer than ( n  1 ) edges, it can’t be a valid tree.
Here’s a simple stepbystep method to determine if a given graph is a valid tree:
 Check the Number of Edges: If the number of edges is not equal to ( n  1 ), return
false
.  Build an Adjacency List: Create an adjacency list from the given edges to represent the graph.
 Perform DepthFirst Search (DFS): Perform a DFS to see if you can visit all the nodes. Keep track of visited nodes to avoid cycles.
 Check if All Nodes Were Visited: After the DFS, if there are any nodes that were not visited, it means the graph is not connected, so return
false
.  If all the above conditions are satisfied, return
true
.
Here’s the code:


Key Insights
 A valid tree must have exactly ( n  1 ) edges.
 A valid tree must be connected, meaning there is a path between every pair of nodes.
 DFS is a convenient way to check if all nodes are connected.
 If any node is not visited after a complete DFS traversal, it means that the graph is not connected, and thus it is not a valid tree.
title: Graph Valid Tree excerpt: Given n nodes labeled from 0 to n1 and a list of undirected edges, check whether these edges make up a valid tree. tags: cycledetection
Given n nodes labeled from 0 to n1 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 1:
Input: n = 5, and edges = [[0,1], [0,2], [0,3], [1,4]]
Output: true
Example 2:
Input: n = 5, and edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]
Output: 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.
Thought Process
Define the Interface
 Input: array of integers (edges), n  number of nodes
 Output: boolean
Identify the Invariants
 Number of nodes will not change
 Edges will also remain the same
 You cannot visit a node more than once
Identify the constraints
 If we have 5 nodes, we can have only n1 edges if it is a graph valid tree
Classify the Problem
 Graph Traversal
Analyze the Examples
Example 1
3

0  1  4

2
We should always have n1 edges to return true. Return true (we have n1 edges for n nodes  no cycle)
Example 2
0  1  2  3

4
Return false. Because we have a cycle in the graph. You have 5 edges and 5 nodes (you have a cycle)
Edge Case
[[0,1], [0,3], [1,4]]
0  1  4

3 2
If you have two components then return false If you cannot reach a node, return false.
We can have more 0 or more neighbors for a given node.
Time and Space Complexity
Time: O(V+E) Space: O(N) + O(N) (for recursion)
Solution Outline
Check for the edge case and return false if it is edge case.
Convert the edges to adjacency list.
Traverse the graph Explore all the neighbors Keep track of visited nodes Detect the cycle
0 1 2 3 4 [t ,t ,f ,f, f]
Return whether the graph has cycle or not
[0] > [1,2,3] [1] > [4, 0] [2] > [0] [3] > [0] [4] > [1]
Key Takeaways
 Break down the problem into smaller subproblems
 Creating the adjacency list from edges
 Graph traversal dfs can be replace with bfs
 Iterative traversal will also work
 Identify the invariant
 Express the invariant in the code to avoid infinite recursion
 Undirected graph You can to create the connection both ways when you create the adjacency list
Implementation


0 1 2 3 4 [t,t,f,f,f]
Define the Interface
 Input: array of integers (edges), n  number of nodes
 Output: boolean
Identify the Invariants
 number of nodes will not change
 edges will also remain the same
 You cannot visit a node more than once
Identify the constraints
 If we have 5 nodes, we can have only n1 edges if it is a graph valid tree
Search the Problem Space
Classify the Problem
 Graph Traversal
Analyze the Examples
Example 1
3  0 — 1 — 4  2
We should always have n1 edges to return true.
Return true (we have n1 edges for n nodes  no cycle)
Example 2
0 — 1 — 2 — 3  4
Return false Because we have a cycle in the graph. You have 5 edges and 5 nodes (you have a cycle)
Edge Case
[[0,1], [0,3], [1,4]]
0 — 1 — 4  3 2
If you have two components then return false If you cannot reach a node, return false.
Solve the Problem by Hand
Describe the Pattern
We can have more 0 or more neighbors for a given node.
Brute Force Approach
Analyze Time and Space Complexity
Time: O( ) Space: O( )
Outline the Solution
 Check for the edge case and return false if it is edge case
 Convert the edges to adjacency list
 Traverse the graph Explore all the neighbors Keep track of visited nodes Detect the cycle
 Return whether the graph has cycle or not
[0] > [1,2,3] [1] > [4, 0] [2] > [0] [3] > [0] [4] > [1]

