Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Check if a Graph is Bipartite Using Depth-First Search (DFS)

Problem Statement

Given an undirected graph represented as an adjacency list with V vertices (0-indexed), determine if the graph is bipartite.

A graph is bipartite if we can split the set of vertices into two groups such that no two adjacent vertices belong to the same group. This is equivalent to checking if the graph can be coloured using two colours without any two adjacent nodes having the same colour.

Examples

Adjacency List Is Bipartite? Description
{ 0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2] }
Yes
Even cycle, can be coloured using two colours
{ 0: [1, 2], 1: [0, 2], 2: [0, 1] }
No
Triangle, odd cycle cannot be bipartite
{}
Yes Empty graph is trivially bipartite
{ 0: [] }
Yes
Single isolated node
{ 0: [1], 1: [0], 2: [3], 3: [2] }
Yes
Two separate components, both bipartite

Visualization Player

Solution

Understanding the Problem

A graph is said to be bipartite if its set of vertices can be divided into two disjoint sets such that every edge connects a vertex from one set to a vertex from the other set. In simpler words, no two adjacent vertices should belong to the same group.

This concept often comes up in real-world scenarios like assigning tasks to two teams such that no conflicting tasks are handled by the same team.

We are given a graph and we have to check if it is bipartite or not. We’ll solve it using Depth-First Search (DFS), by trying to color the graph using two colors and checking for any coloring conflicts.

Step-by-Step Solution with Example

Step 1: Understand the Input Format

The graph is typically represented as an adjacency list. Each key in the list is a node, and its value is an array of nodes it is directly connected to. Example:

{
  0: [1, 3],
  1: [0, 2],
  2: [1, 3],
  3: [0, 2]
}

This graph has 4 nodes (0 to 3) and edges between them. We will now try to determine if this graph can be divided into two groups as per bipartite rules.

Step 2: Initialize Color Map

We will use a map or array to keep track of the color assigned to each node. Let’s say color 0 for one group, and 1 for the other.

Initially, no node is colored.

Step 3: DFS Traversal with Coloring

We start DFS traversal from any unvisited node (say node 0), assign it color 0, and then try to assign its neighbors the opposite color (1).

If at any point, a neighbor already has the same color as the current node, we conclude the graph is not bipartite.

Step 4: Continue for All Components

Since the graph may be disconnected, we need to repeat the DFS traversal from every unvisited node to ensure all components are checked.

Step 5: Example Walkthrough

Let’s walk through the example:

{
  0: [1, 3],
  1: [0, 2],
  2: [1, 3],
  3: [0, 2]
}

Start from node 0 → color it 0

→ visit node 1 → color it 1

→ visit node 2 from node 1 → color it 0

→ visit node 3 from node 2 → color it 1

→ back to node 0, already colored 0 → no conflict

→ visit node 2 from node 3 → already colored 0 → no conflict

All nodes visited without conflict. So this graph is bipartite.

Edge Cases

Case 1: Disconnected Graph

Always check all nodes, not just the first component. Use a loop to start DFS from each unvisited node.

Case 2: Odd-Length Cycle

If the graph contains an odd-length cycle, it cannot be bipartite. For example:

{
  0: [1, 2],
  1: [0, 2],
  2: [0, 1]
}

This is a triangle (3-cycle). You cannot color it with two colors without conflict. Hence, not bipartite.

Case 3: Empty Graph

A graph with no edges is trivially bipartite since no adjacent nodes exist to cause conflicts.

Case 4: Self-Loop

If a node has an edge to itself, the graph is automatically not bipartite, since a node would need to have two different colors.

Finally

Checking whether a graph is bipartite using DFS is a classic and intuitive algorithm. By coloring nodes alternately and validating the coloring rules, we can confidently determine the bipartiteness of a graph.

Always remember to handle disconnected components and edge cases like odd cycles and self-loops to ensure your solution is robust.

Algorithm Steps

  1. Create a color array of size V initialized with -1 (unvisited).
  2. Define a recursive dfs function that takes current node and currentColor.
  3. Color the node and for each of its neighbors:
    1. If neighbor is not coloured, call dfs with opposite colour.
    2. If already coloured and same as current node, return false.
  4. If DFS completes without conflicts, return true.
  5. Check each unvisited node to ensure disconnected components are handled.

Code

JavaScript
function isBipartiteDFS(adj, V) {
  const color = new Array(V).fill(-1);

  function dfs(node, c) {
    color[node] = c;

    for (const neighbor of adj[node] || []) {
      if (color[neighbor] === -1) {
        if (!dfs(neighbor, 1 - c)) return false;
      } else if (color[neighbor] === color[node]) {
        return false;
      }
    }
    return true;
  }

  for (let i = 0; i < V; i++) {
    if (color[i] === -1) {
      if (!dfs(i, 0)) return false;
    }
  }
  return true;
}

// Example usage
const graph1 = { 0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2] };
console.log("Is graph1 bipartite?", isBipartiteDFS(graph1, 4)); // true

const graph2 = { 0: [1, 2], 1: [0, 2], 2: [0, 1] };
console.log("Is graph2 bipartite?", isBipartiteDFS(graph2, 3)); // false

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(V + E)Even in the best case, we must traverse all vertices and their edges to ensure bipartite coloring without assumptions.
Average CaseO(V + E)Each vertex and edge is visited once during the DFS traversal.
Worst CaseO(V + E)In the worst case, the graph is fully connected and requires full traversal of all vertices and edges.

Space Complexity

O(V)

Explanation: We use a color array of size V and recursive call stack of maximum depth V in the worst case.


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