Tutorialspoint.dev

Level of Each node in a Tree from source node (using BFS)

BFS(Breadth First Search) is a graph traversal technique where a node and its neighbors are visited first and then the neighbors of neighbors. In simple terms it traverses level wise from the source. First it traverses level 1 nodes (direct neighbors of source node) and then level 2 nodes (neighbors of neighbors of source node) and so on.

Now, suppose if we have to know at which level all the nodes are at (from source node). Then BFS can be used to determine the level of each node.

Examples:

Input :   

Output :  Node      Level
           0          0
           1          1
           2          1
           3          2
           4          2
           5          2
           6          2
           7          3

Explanation : 

C++

// CPP Program to determine level of each node
// and print level
#include <iostream>
#include <queue>
#include <vector>
using namespace std;
  
// function to determine level of each node starting
// from x using BFS
void printLevels(vector<int> graph[], int V, int x)
{
    // array to store level of each node
    int level[V];
    bool marked[V];
  
    // create a queue
    queue<int> que;
  
    // enqueue element x
    que.push(x);
  
    // initialize level of source node to 0
    level[x] = 0;
  
    // marked it as visited
    marked[x] = true;
  
    // do until queue is empty
    while (!que.empty()) {
  
        // get the first element of queue
        x = que.front();
  
        // dequeue element
        que.pop();
  
        // traverse neighbors of node x
        for (int i = 0; i < graph[x].size(); i++) {
            // b is neighbor of node x
            int b = graph[x][i];
  
            // if b is not marked already
            if (!marked[b]) {
  
                // enqueue b in queue
                que.push(b);
  
                // level of b is level of x + 1
                level[b] = level[x] + 1;
  
                // mark b
                marked[b] = true;
            }
        }
    }
  
    // display all nodes and their levels
    cout << "Nodes"
         << "    "
         << "Level" << endl;
    for (int i = 0; i < V; i++)
        cout << " " << i << "   -->   " << level[i] << endl;
}
  
// Dirver Code
int main()
{
    // adjacency graph for tree
    int V = 8;
    vector<int> graph[V];
  
    graph[0].push_back(1);
    graph[0].push_back(2);
    graph[1].push_back(3);
    graph[1].push_back(4);
    graph[1].push_back(5);
    graph[2].push_back(5);
    graph[2].push_back(6);
    graph[6].push_back(7);
  
    // call levels function with source as 0
    printLevels(graph, V, 0);
  
    return 0;
}

Python3

# Python3 Program to determine level
# of each node and print level
import queue

# function to determine level of
# each node starting from x using BFS
def printLevels(graph, V, x):



# array to store level of each node
level = [None] * V
marked = [False] * V

# create a queue
que = queue.Queue()

# enqueue element x
que.put(x)

# initialize level of source
# node to 0
level[x] = 0

# marked it as visited
marked[x] = True

# do until queue is empty
while (not que.empty()):

# get the first element of queue
x = que.get()

# traverse neighbors of node x
for i in range(len(graph[x])):

# b is neighbor of node x
b = graph[x][i]

# if b is not marked already
if (not marked[b]):



# enqueue b in queue
que.put(b)

# level of b is level of x + 1
level[b] = level[x] + 1

# mark b
marked[b] = True

# display all nodes and their levels
print(“Nodes”, ” “, “Level”)
for i in range(V):
print(” “,i, ” –> “, level[i])

# Driver Code
if __name__ == ‘__main__’:

# adjacency graph for tree
V = 8
graph = [[] for i in range(V)]

graph[0].append(1)
graph[0].append(2)
graph[1].append(3)
graph[1].append(4)
graph[1].append(5)
graph[2].append(5)
graph[2].append(6)
graph[6].append(7)

# call levels function with source as 0
printLevels(graph, V, 0)

# This code is contributed by PranchalK


Output:

Nodes    Level
 0   -->   0
 1   -->   1
 2   -->   1
 3   -->   2
 4   -->   2
 5   -->   2
 6   -->   2
 7   -->   3


This article is attributed to GeeksforGeeks.org

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter