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

13 Aug, 2021
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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to calculate the degree of all the verticesvoid 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 queriesvoid 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 codeint 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 implementation of the approachimport 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 verticesstatic 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 queriesstatic 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 codepublic 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 implementation of the approach # Function to calculate the degree# of all the verticesdef 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 queriesdef 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 codeif __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# implementation of the approachusing 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 verticesstatic 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 queriesstatic 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 codepublic 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
<script> // JavaScript implementation of the approach // Function to calculate the degree of all the verticesfunction 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 queriesfunction 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 verticesvar n = 6;// Edges of the treevar edges = [    [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 1, 4 ], [ 4, 5 ]];// Queriesvar q = [ 0, 3, 4, 5 ];// Perform the queriesperformQueries(edges, q, n); </script>
No
Yes
No
Yes
 
Time complexity: O(n)Auxiliary Space: O(n).  
Arrays
Competitive Programming
Data Structures
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