What is the optimal approach and complexity for Graph Valid Tree?

Need exactly n − 1 edges, and no union may fail with optimal time complexity O(E · α(N)) and space complexity O(N).

How do you identify the pattern for Graph Valid Tree?

Look for key signals in the prompt: "valid tree" indicates Connected and no cycle, at the same time → count the edges and watch for a failed union.; "undirected edges" indicates Direction does not matter, so one union per edge is enough.; "n nodes and a list of edges" indicates The count n − 1 is a free check before any union happens..

Graph Valid Tree

Medium·LeetCode #261 Premium Pro OnlySlight Variation

LeetCode

Problem Statement

You have n nodes labelled 0 to n − 1 and a list of undirected edges, where edges[i] = [a, b] is one link between nodes a and b.

Return true if these edges form a valid tree, and false otherwise.

A valid tree is a graph that is connected (you can get from any node to any other node) and has no cycle (no way to leave a node and come back to it without repeating an edge).

Input

n, the number of nodes, and edges, a list of [a, b] pairs.

Output

true if the edges form a tree, otherwise false.

Example 1

Input: n = 5, edges = [[0,1], [0,2], [0,3], [1,4]]

Output: true

Node 0 is the hub: nodes 1, 2 and 3 hang off it, and node 4 hangs off node 1. Five nodes, four edges, and no cycle.

Example 2

Input: n = 5, edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]

Output: false

The links 1–2, 2–3 and 3–1 form a cycle. Every node is connected, but a tree cannot have a cycle.

Constraints

• Nodes are labelled 0 to n − 1.
• Each edge is a pair [a, b] of node labels from 0 to n − 1.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“valid tree”

→ Connected and no cycle, at the same time → count the edges and watch for a failed union.

“undirected edges”

→ Direction does not matter, so one union per edge is enough.

“n nodes and a list of edges”

→ The count n − 1 is a free check before any union happens.

Approach Complexity Ladder

Brute Force — BFS for each edge, then one BFS from node 0O(E · (N + E))O(N + E)

Each edge: BFS over the earlier edges to see if its ends are joined; then one BFS to check connectivity

Optimal — count the edges, then unite themO(E · α(N))O(N)

Need exactly n − 1 edges, and no union may fail

Brute Force Intuition

For each edge (a, b), run a BFS from a over the earlier edges only. If b is reached, the edge closes a cycle, so the answer is false. If no edge does, run one BFS from node 0 over all edges: a tree must reach all n nodes.

Recognition rule for this sub-pattern:

Start with n groups; each successful union reduces the count by one.

Remember this · Union-Find (DSU)

Trigger

Edges arrive one by one, "are these connected?", merge equivalent items, count components, extra edge creating a cycle.

Think

Am I repeatedly merging groups and asking whether two things are in the same group?

Key question

What are the elements, and what does it mean to union two of them?

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