Check whether a node is leaf node or not for multiple queries

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Given a tree with N vertices numbered from 0 to N – 1 where 0 is the root node. The task is to check if a node is leaf node or not for multiple queries.
Examples:

Input:
       0
     /   \
   1      2
 /  \
3    4 
    /
  5
q[] = {0, 3, 4, 5}
Output:
No
Yes
No
Yes
From the graph, 2, 3 and 5 are the only leaf nodes.

Approach: Store the degree of all the vertices in an array degree[]. For each edge from A to B, degree[A] and degree[B] are incremented by 1. Now every node which not a root node and it has a degree of 1 is a leaf node and all the other nodes are not.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the degree of all the vertices
void init(int degree[], vector<pair<int, int> > edges, int n)
{
    // Initializing degree of all the vertices as 0
    for (int i = 0; i < n; i++) {
        degree[i] = 0;
    }
 
    // For each edge from A to B, degree[A] and degree[B]
    // are increased by 1
    for (int i = 0; i < edges.size(); i++) {
        degree[edges[i].first]++;
        degree[edges[i].second]++;
    }
}
 
// Function to perform the queries
void performQueries(vector<pair<int, int> > edges,
                    vector<int> q, int n)
{
    // To store the of degree
    // of all the vertices
    int degree[n];
 
    // Calculate the degree for all the vertices
    init(degree, edges, n);
 
    // For every query
    for (int i = 0; i < q.size(); i++) {
 
        int node = q[i];
        if (node == 0) {
            cout << "No\n";
            continue;
        }
        // If the current node has 1 degree
        if (degree[node] == 1)
            cout << "Yes\n";
        else
            cout << "No\n";
    }
}
 
// Driver code
int main()
{
 
    // Number of vertices
    int n = 6;
 
    // Edges of the tree
    vector<pair<int, int> > edges = {
        { 0, 1 }, { 0, 2 }, { 1, 3 }, { 1, 4 }, { 4, 5 }
    };
 
    // Queries
    vector<int> q = { 0, 3, 4, 5 };
 
    // Perform the queries
    performQueries(edges, q, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG 
{
static class pair
{ 
    int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
} 
 
// Function to calculate the degree 
// of all the vertices
static void init(int degree[],
                     pair[] edges, int n)
{
    // Initializing degree of 
    // all the vertices as 0
    for (int i = 0; i < n; i++) 
    {
        degree[i] = 0;
    }
 
    // For each edge from A to B, 
    // degree[A] and degree[B]
    // are increased by 1
    for (int i = 0; i < edges.length; i++) 
    {
        degree[edges[i].first]++;
        degree[edges[i].second]++;
    }
}
 
// Function to perform the queries
static void performQueries(pair [] edges,
                           int []q, int n)
{
    // To store the of degree
    // of all the vertices
    int []degree = new int[n];
 
    // Calculate the degree for all the vertices
    init(degree, edges, n);
 
    // For every query
    for (int i = 0; i < q.length; i++)
    {
 
        int node = q[i];
        if (node == 0)
        {
            System.out.println("No");
            continue;
        }
         
        // If the current node has 1 degree
        if (degree[node] == 1)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// Driver code
public static void main(String[] args)
{
    // Number of vertices
    int n = 6;
 
    // Edges of the tree
    pair[] edges = {new pair(0, 1), 
                    new pair(0, 2),
                    new pair(1, 3), 
                    new pair(1, 4), 
                    new pair(4, 5)};
 
    // Queries
    int []q = { 0, 3, 4, 5 };
 
    // Perform the queries
    performQueries(edges, q, n);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
 
# Function to calculate the degree
# of all the vertices 
def init(degree, edges, n) : 
 
    # Initializing degree of
    # all the vertices as 0 
    for i in range(n) :
        degree[i] = 0; 
 
    # For each edge from A to B, 
    # degree[A] and degree[B] 
    # are increased by 1 
    for i in range(len(edges)) :
        degree[edges[i][0]] += 1; 
        degree[edges[i][1]] += 1; 
 
# Function to perform the queries 
def performQueries(edges, q, n) : 
 
    # To store the of degree 
    # of all the vertices 
    degree = [0] * n; 
 
    # Calculate the degree for all the vertices 
    init(degree, edges, n); 
 
    # For every query 
    for i in range(len(q)) :
 
        node = q[i]; 
        if (node == 0) :
            print("No"); 
            continue; 
 
        # If the current node has 1 degree 
        if (degree[node] == 1) :
            print("Yes"); 
        else :
            print("No"); 
 
# Driver code 
if __name__ == "__main__" : 
 
    # Number of vertices 
    n = 6; 
 
    # Edges of the tree 
    edges = [[ 0, 1 ], [ 0, 2 ], 
             [ 1, 3 ], [ 1, 4 ], 
             [ 4, 5 ]]; 
 
    # Queries 
    q = [ 0, 3, 4, 5 ]; 
 
    # Perform the queries 
    performQueries(edges, q, n); 
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
                     
class GFG 
{
public class pair
{ 
    public int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
} 
 
// Function to calculate the degree 
// of all the vertices
static void init(int []degree,
                 pair[] edges, int n)
{
    // Initializing degree of 
    // all the vertices as 0
    for (int i = 0; i < n; i++) 
    {
        degree[i] = 0;
    }
 
    // For each edge from A to B, 
    // degree[A] and degree[B]
    // are increased by 1
    for (int i = 0; i < edges.Length; i++) 
    {
        degree[edges[i].first]++;
        degree[edges[i].second]++;
    }
}
 
// Function to perform the queries
static void performQueries(pair [] edges,
                            int []q, int n)
{
    // To store the of degree
    // of all the vertices
    int []degree = new int[n];
 
    // Calculate the degree for all the vertices
    init(degree, edges, n);
 
    // For every query
    for (int i = 0; i < q.Length; i++)
    {
 
        int node = q[i];
        if (node == 0)
        {
            Console.WriteLine("No");
            continue;
        }
         
        // If the current node has 1 degree
        if (degree[node] == 1)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// Driver code
public static void Main(String[] args)
{
    // Number of vertices
    int n = 6;
 
    // Edges of the tree
    pair[] edges = {new pair(0, 1), 
                    new pair(0, 2),
                    new pair(1, 3), 
                    new pair(1, 4), 
                    new pair(4, 5)};
 
    // Queries
    int []q = { 0, 3, 4, 5 };
 
    // Perform the queries
    performQueries(edges, q, n);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// JavaScript implementation of the approach
 
// Function to calculate the degree of all the vertices
function init(degree, edges, n)
{
    // Initializing degree of all the vertices as 0
    for (var i = 0; i < n; i++) {
        degree[i] = 0;
    }
 
    // For each edge from A to B, degree[A] and degree[B]
    // are increased by 1
    for (var i = 0; i < edges.length; i++) {
        degree[edges[i][0]]++;
        degree[edges[i][1]]++;
    }
}
 
// Function to perform the queries
function performQueries( edges, q, n)
{
    // To store the of degree
    // of all the vertices
    var degree = Array(n);
 
    // Calculate the degree for all the vertices
    init(degree, edges, n);
 
    // For every query
    for (var i = 0; i < q.length; i++) {
 
        var node = q[i];
        if (node == 0) {
            document.write( "No<br>");
            continue;
        }
        // If the current node has 1 degree
        if (degree[node] == 1)
            document.write( "Yes<br>");
        else
            document.write( "No<br>");
    }
}
 
// Driver code
// Number of vertices
var n = 6;
// Edges of the tree
var edges = [
    [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 1, 4 ], [ 4, 5 ]
];
// Queries
var q = [ 0, 3, 4, 5 ];
// Perform the queries
performQueries(edges, q, n);
 
</script>

Output:

No
Yes
No
Yes

Time complexity: O(n)
Auxiliary Space: O(n).

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