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

Last Updated : 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++

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

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