A **graph** is a data structure that consists of a set of nodes connected by edges. Graphs are used to simulate many real-world problems, such as paths in cities, circuit networks, and social networks. This is graph implementation part 1 – unweighted graph.

### How to represent nodes and edges in adjacency list?

An adjacency list is a list of list: a list of nodes, with each node has a list of neighbors. The underneath data structure is a hash map. The buckets of hash map hold all nodes(vertices) in a graph. The node is the key. The value is the linked list of the node’s neighbors.

### Table of Content

- Terminology
- Map of graph implementations
- Add node and edge
- Remove node and edge
- Search by node and edge
- Find path with DFS in graph
- Find path with BFS in graph
- print graph as adjacency list
- Traverse with DFS in graph
- Traverse with BFS in graph
- Free download

Terminology

A node in the graph is also called **vertex**. A line between two nodes is **edge**. Two nodes are **adjacent** (or neighbors) if they are connected to each other through an edge.** Path** represents a sequence of edges between the two nodes.

In a **directed** graph, all of the edges represent a one-way relationship.** **In an **undirected **graph, all edges are bi-directional.

If the edges in the graph have weights (represent costs or distances), the graph is said to be a **weighted** graph. If the edges do not have weights, the graph is said to be** unweighted**.

Graph can be presented as adjacency list or adjacency matrix. An **adjacency list** is an array of edges or nodes**.** Adjacency list is used for representation of the sparse graphs and used more often. An **adjacency matrix **is a square matrix with dimensions equivalent to the number of nodes in the graph. Adjacency matrix is preferred when the graph is dense.

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Part 1 – Graph implementation as adjacency list

Part 2 – Weighted graph as adjacency list

Part 3 – Graph as adjacency matrix

Add node and edge

The Graph class has two fields: *adj* and *directed*. *adj *is a HashMap in which the key is the node, the value is all its neighbors. The value is represented as linked list of the nodes. *directed* is a boolean variable to specify whether the graph is directed or undirected. By default, it is undirected.

In many old implementation using adjacency list, *adj *is defined as an array of node list. The disadvantage of this implementation is that we have to define the length of the array first. To avoid this, we use hashmap. By using hashmap, we can retrieve node and its neighbors by node itself, not by index in array. Note the node can be any data type, for example primitive data type, such as integer or string, or an object, such as graphNode.

To add a node to the graph is to add a key in the hashmap. To add an edge is to add an item in this key’s value. The following method *addEdge* includes both adding a node and adding an edge. For a directed graph, we add edge from *a* to *b*. For undirected graph, we also add edge from *b* to *a*.

## Java

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public class Graph<T> { Map<T, LinkedList<T>> adj = new HashMap<>(); boolean directed; //Constructor, Time O(1) Space O(1) public Graph() { directed = false; //default, Undirected graph } //Constructor, Time O(1) Space O(1) public Graph(boolean d) { directed = d; } //Add edges including adding nodes, Time O(1) Space O(1) public void addEdge(T a, T b) { adj.putIfAbsent(a, new LinkedList<>()); //add node adj.putIfAbsent(b, new LinkedList<>()); //add node adj.get(a).add(b); //add edge if (!directed) { //undirected adj.get(b).add(a); } } } |

## Javascript

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class Graph { //Constructor, Time O(1) Space O(1) constructor(directed) { this.adj = new Map(); this.directed = directed; //true or false } //Add edges including adding nodes, Time O(1) Space O(1) addEdge(a, b) { if (this.adj.get(a) == null) this.adj.set(a, new Array()); if (this.adj.get(b) == null) this.adj.set(b, new Array()); this.adj.get(a).push(b); if (!this.directed) this.adj.get(b).push(a); } } |

## Python

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class Graph : #Constructor, Time O(1) Space O(1) def __init__(self, directed) : self.adj = {} self.directed = directed #true or false #Add edges including adding nodes, Time O(1) Space O(1) def addEdge(self, a, b) : if a not in self.adj: self.adj[a] = [] if b not in self.adj: self.adj[b] = [] self.adj[a].append(b) if (self.directed == False) : self.adj[b].append(a) |

## Doodle

Remove node and edge

Remove operation includes remove edge and remove node. To **remove edge**, we use the node as key to find its neighbors in the hashmap. Then remove the other node from its neighbors. For a directed graph, we just need to remove edge from *a* to *b*. For an undirected graph, we also need to remove the edge from *b* to *a*.

## Java

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//Remove direct connection between a and b, Time O(1) Space O(1) public void removeEdge(T a, T b) { LinkedList<T> ne1 = adj.get(a); LinkedList<T> ne2 = adj.get(b); if (ne1 == null || ne2 == null) return; ne1.remove(b); if (!directed) //undirected ne2.remove(a); } |

## Javascript

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//Remove direct connection between a and b, Time O(1) Space O(1) removeEdge(a, b) { var ne1 = this.adj.get(a); var ne2 = this.adj.get(b); if (ne1 == null || ne2 == null) return; var index = ne1.indexOf(b); if (index >= 0) ne1.splice(index, 1); if (!this.directed) { index = ne2.indexOf(a); ne2.splice(index, 1) } } |

## Python

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def removeEdge(self, a, b) : ne1 = self.adj[a] ne2 = self.adj[b] if ne1 == None or ne2 == None : return ne1.remove(b) if (self.directed == False) : ne2.remove(a) |

## Doodle

**Remove node** has more work to do than remove edge. We have to remove all connected edge before remove the node itself. For an undirected graph, first we get all neighbors of the node. Then for each of its neighbors, remove itself from the value list. For a directed graph, we search all keys in the hashmap for their values, and check whether this node exists in their neighbors. If it does, remove it. The last step is to remove the node as the key in the hashmap. Then this node is no longer in the hashmap’s key set.

## Java

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//Remove a node including all its edges, //Time O(n) Space O(1), n is number of its neighbors public void removeNode(T a) { if (!directed) { //undirected LinkedList<T> ne1 = adj.get(a); for (T node: ne1) { adj.get(node).remove(a); } } else { //directed for (T key: adj.keySet()) { adj.get(key).remove(a); } } adj.remove(a); } |

## Javascript

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//Remove a node including all its edges, //Time O(n) Space O(1), n is number of its neighbors removeNode(a) { if (!this.directed) { //undirected var ne1 = this.adj.get(a); for (let node of ne1) { let list = this.adj.get(node); let index = list.indexOf(a); if (index >= 0) list.splice(index, 1); } } else { //directed for (let entry of this.adj.entries()) { let list = entry[1]; let index = list.indexOf(a); if (index >= 0) list.splice(index, 1); } } this.adj.delete(a); } |

## Python

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#Remove a node including all its edges, #Time O(n) Space O(1), n is number of its neighbors def removeNode(self, a) : if self.directed == False: #undirected ne1 = self.adj[a] for node in ne1 : self.adj[node].remove(a) else : #directed for k, v in self.adj.items(): if a in v: v.remove(a); self.adj.pop(a) |

## Doodle

Search by node and edge

Search can be search node, edge or path. We can check whether there is a node existing in the graph. This can be done by simply checking the hashmap contains the key. We can also check whether there is a direct connection between two nodes (aka whether there is an edge). This can be done by checking whether the other node is in one node’s neighbors.

## Java

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//Check whether there is node by its key, Time O(1) Space O(1) public boolean hasNode(T key) { return adj.containsKey(key); } //Check whether there is direct connection between two nodes, Time O(1), Space O(1) public boolean hasEdge(T a, T b) { LinkedList<T> ne1 = adj.get(a); if (directed) //directed return ne1.contains(b); else { //undirected or bi-directed LinkedList<T> ne2 = adj.get(b); return ne1.contains(b) && ne2.contains(a); } } |

## Javascript

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//Check whether there is node with the key, Time O(1) Space O(1) hasNode(key) { return this.adj.has(key); } //Check whether there is direct connection between two nodes, Time O(1), Space O(1) hasEdge(a, b) { var ne1 = this.adj.get(a); if (this.directed) //directed return ne1.includes(b); else { //undirected or bi-directed var ne2 = this.adj.get(b); return ne1.includes(b) && ne2.includes(a); } } |

## Python

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#Check whether there is node by its key, Time O(1) Space O(1) def hasNode(self, key) : return key in self.adj.keys() #Check whether there is direct connection between two nodes, Time O(1), Space O(1) def hasEdge(self, a, b): ne1 = [] ne2 = [] if a in self.adj: ne1 = self.adj[a] if self.directed : #directed return b in ne1 else : #undirected or bi-directed if b in self.adj: ne2 = self.adj[b] return b in ne1 and a in ne2 |

## Doodle

Find path with depth first search in graph

More useful operation is to search path. A path is a sequence of edges. There can be more than one path between two nodes. There are two common approaches: **depth first search** (DFS) and **breadth first search** (BFS). In this section, we use DFS and BFS to find out whether there is path from one node to another. In traversal section, we use them to find all reachable nodes from the source node in graph. DFS and BFS is most importation part of graph implementation.

**Depth First Search **starts from the source node, and explores the adjacent nodes as far as possible before call back. DFS is usually implemented with recursion or stack. It is used to solve find path or detect cycle problems.

## Java

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//Check there is path from src and dest //DFS, Time O(V+E), Space O(V) public boolean dfs(T src, T dest) { HashMap<T, Boolean> visited = new HashMap<>(); return dfsHelper(src, dest, visited); } //DFS helper, Time O(V+E), Space O(V) private boolean dfsHelper(T v, T dest, HashMap<T, Boolean> visited) { if (v == dest) return true; visited.put(v, true); for (T ne : adj.get(v)) { if (visited.get(ne) == null) return dfsHelper(ne, dest, visited); } return false; } |

## Javascript

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//Check there is path from src and dest //DFS, Time O(V+E), Space O(V) dfs(src, dest) { var visited = new Map(); return this.dfsHelper(src, dest, visited); } //DFS helper, Time O(V+E), Space O(V) dfsHelper(v, dest, visited) { if (v == dest) return true; visited.set(v, true); for (let ne of this.adj.get(v)) { if (visited.get(ne) == null) return this.dfsHelper(ne, dest, visited); } return false; } |

## Python

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# Check there is path from src and dest # DFS, Time O(V+E), Space O(V) def dfs(self, src, dest) : visited = {} return self.dfsHelper(src, dest, visited) #DFS helper, Time O(V+E), Space O(V) def dfsHelper(self, v, dest, visited) : if v == dest: return True visited[v] = True for ne in self.adj[v] : if ne not in visited: return self.dfsHelper(ne, dest, visited) return False |

## Doodle

Find path with breadth first search in graph

**Breath First Search ** starts from the source node, and explores all its adjacent nodes before going to the next level adjacent nodes. BFS is usually implemented with Queue. It is often used to solve shortest path problems.

## Java

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//Check there is path from src and dest, may include multiple edges //BFS, Time O(V+E), Space O(V), V is number of vertices, E is number of edges public boolean bfs(T src, T dest) { HashMap<T, Boolean> visited = new HashMap<>(); Queue<T> q = new LinkedList<>(); visited.put(src, true); q.add(src); while (!q.isEmpty()) { T v = q.poll(); for (T ne : adj.get(v)) { if (ne == dest) return true; if (visited.get(ne) == null) { q.add(ne); visited.put(ne, true); } } } return false; } |

## Javascript

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//Check there is path from src and dest, may include multiple edges //BFS, Time O(V+E), Space O(V), V is number of vertices, E is number of edges bfs(src, dest) { var visited = new Map(); var q = new Array(); visited.set(src, true); q.push(src); while (q.length > 0) { let v = q.shift(); for (let ne of this.adj.get(v)) { if (ne == dest) return true; if (visited.get(ne) == null) { q.push(ne); visited.set(ne, true); } } } return false; } |

## Python

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#Check there is path from src and dest, may include multiple edges #BFS, Time O(V+E), Space O(V), V is number of vertices, E is number of edges def bfs(self, src, dest) : visited = {} q = [] visited[src] = True q.append(src) while q : v = q.pop(0); for ne in self.adj[v] : if ne == dest: return True if ne not in visited: q.append(ne) visited[ne] = True return False |

Print graph as adjacency list

Print is to visit all nodes in the graph and print the information stored. This is actually to the hashmap. It can be done by looping through the key set of the hashmap. This method is used for debugging purpose.

## Java

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//Print graph as hashmap, Time O(V+E), Space O(1) public void print() { for (T key: adj.keySet()) { System.out.println(key + "," + adj.get(key)); } } |

## JavaScript

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//Print graph as hashmap, Time O(V+E), Space O(1) print() { for (let entry of this.adj.entries()) { console.log(entry[0] + "-" + entry[1]); } } |

## Python

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# Print graph as hashmap, Time O(V+E), Space O(1) def printGraph(self) : for k, v in self.adj.items(): print(str(k) + "-" + str(v)) |

Traverse with depth first search in graph

**DFS traversal**: Use depth first search to visit nodes in the graph and print the node’s information. This is similar to DFS traversal in binary tree. Starting from the source node, we call recursive method to visit its neighbor’s neighbor until call back. Please node the source might be any node in the graph. Some nodes might not be reached in a directed graph. (In binary tree, we always start from the root and all nodes should be visited.)

## Java

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//Traversal starting from src, DFS, Time O(V+E), Space O(V) public void dfsTraversal(T src) { HashMap<T, Boolean> visited = new HashMap<>(); helper(src, visited); } //DFS helper, Time O(V+E), Space O(V) private void helper(T v, HashMap<T, Boolean> visited) { visited.put(v, true); System.out.print(v.toString() + " "); for (T ne : adj.get(v)) { if (visited.get(ne) == null) helper(ne, visited); } } |

## JavaScript

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//Traversal starting from src, DFS, Time O(V+E), Space O(V) dfsTraversal(src) { var visited = new Map(); this.helper(src, visited); } //DFS helper, Time O(V+E), Space O(V) helper(v, visited) { visited.set(v, true); console.log(v.toString()); for (let ne of this.adj.get(v)) { if (visited.get(ne) == null) this.helper(ne, visited); } } |

## Python

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#Traversal starting from src, DFS, Time O(V+E), Space O(V) def dfsTraversal(self, src) : visited = {} self.helper(src, visited) #DFS helper, Time O(V+E), Space O(V) def helper(self, v, visited) : visited[v] = True print(str(v)) for ne in self.adj[v] : if ne not in visited: self.helper(ne, visited) |

## Doodle

Traverse with breadth first search in graph

**BFS traversal**: Use breadth first search to visit all nodes in the graph and print the node’s information. This is similar to BFS traversal in binary tree. Starting from the source, visit all its neighbors first before visiting neighbor’s neighbor. Please node the source might be any node in the graph. Some nodes might not be reached in a directed graph. (In binary tree, we always start from the root and all nodes should be visited.)

## Java

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//Traversal starting from src, BFS, Time O(V+E), Space O(V) public void bfsTraversal(T src) { Queue<T> q = new LinkedList<>(); HashMap<T, Boolean> visited = new HashMap<>(); q.add(src); visited.put(src, true); while (!q.isEmpty()) { T v = q.poll(); System.out.print(v.toString() + " "); for (T ne : adj.get(v)) { if (visited.get(ne) == null) { q.add(ne); visited.put(ne, true); } } } System.out.println(); } |

## JavaScript

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//Traversal starting from src, BFS, Time O(V+E), Space O(V) bfsTraversal(src) { var q = new Array(); var visited = new Map(); q.push(src); visited.set(src, true); while (q.length > 0) { let v = q.shift(); console.log(v.toString() + " "); for (let ne of this.adj.get(v)) { if (visited.get(ne) == null) { q.push(ne); visited.set(ne, true); } } } } |

## Python

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# Traversal starting from src, BFS, Time O(V+E), Space O(V) def bfsTraversal(self, src) : q = [] visited = {} q.append(src) visited[src] = True while (len(q) > 0) : v = q.pop(0) print(str(v) + " "); for ne in self.adj[v] : if ne not in visited: q.append(ne); visited[ne] = True |

## Doodle

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