Leetcode #261: Graph Valid Tree

In this guide, we solve Leetcode #261 Graph Valid Tree in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You have a graph of n nodes labeled from 0 to n - 1. You are given an integer n and a list of edges where edges[i] = [ai, bi] indicates that there is an undirected edge between nodes ai and bi in the graph.

Quick Facts

Difficulty: Medium
Premium: Yes
Tags: Depth-First Search, Breadth-First Search, Union Find, Graph

Intuition

The data forms a graph, so we should explore nodes and edges systematically.

A traversal ensures we visit each node once while maintaining the needed state.

Approach

Build an adjacency list and traverse with BFS or DFS.

Aggregate results as you visit nodes.

Steps:

Build the graph.
Traverse with BFS/DFS.
Accumulate the required output.

Example

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

Python Solution

class Solution:
    def validTree(self, n: int, edges: List[List[int]]) -> bool:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(n))
        for a, b in edges:
            pa, pb = find(a), find(b)
            if pa == pb:
                return False
            p[pa] = pb
            n -= 1
        return n == 1

Complexity

The time complexity is $O(n \times \log n)$ , and the space complexity is $O(n)$ , where $n$ is the number of nodes. The space complexity is $O(n)$ , where $n$ is the number of nodes.

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Python Solution

class Solution:
    def validTree(self, n: int, edges: List[List[int]]) -> bool:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(n))
        for a, b in edges:
            pa, pb = find(a), find(b)
            if pa == pb:
                return False
            p[pa] = pb
            n -= 1
        return n == 1

Leetcode #261: Graph Valid Tree

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #261: Graph Valid Tree

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview