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Number of special nodes in an n-ary tree
  • Difficulty Level : Medium
  • Last Updated : 02 Jun, 2021

Given an n-ary tree rooted at vertex 1. The tree has n vertices and n-1 edges. Each node has a value associated with it and tree is input in the form of adjacency list. The task is to find the number of special nodes in the tree. A node is special if the path from the root to the node consists of distinct value nodes.
Examples: 
 

Input: val[] = {1, 2, 3, 4, 5, 7, 2, 3}
                1
               / \
              2   3
             / \   \
            4   5   7
               / \
              2   3
     
Output: 7
All nodes except the leaf node 2 are special.

Input: val[] = {2, 1, 4, 3, 4, 8, 10, 2, 5, 1}
                2
               / \
              1   4
            / \ \  \
           3  4  8  10
            / \ \
           2  5  1
    
Output: 8
Leaf nodes 2 and 1 are not special.

 

Approach: The idea is to perform dfs on given tree using adjacency list. While performing dfs insert values of nodes visited in a set. If value of current node is already present in the set then current node and all nodes in the subtree rooted at current node are not special. After traversing subtree rooted at current node erase the value of current node from set as this value or node does not lie on path from root to all other unvisited nodes. 
Below is the implementation of the above approach:
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// DFS function to traverse the tree and find
// number of special nodes
void dfs(int val[], int n, vector<int> adj[], int v,
         unordered_set<int>& values, int& ans)
{
 
    // If value of current node is already
    // present earlier in path then this
    // node and all other nodes connected to
    // it are not special
    if (values.count(val[v]))
        return;
 
    // Insert value of current node in
    // set of values traversed
    ans++;
    values.insert(val[v]);
 
    // Call dfs on all adjacent nodes
    for (auto ele : adj[v]) {
        dfs(val, n, adj, ele, values, ans);
    }
 
    // Erase value of current node as all paths
    // passing through current node are traversed
    values.erase(val[v]);
}
 
// Driver code
int main()
{
    int val[] = { 0, 2, 1, 4, 3, 4, 8, 10, 2, 5, 1 };
    int n = sizeof(val) / sizeof(val[0]);
 
    vector<int> adj[n];
 
    adj[1].push_back(2);
    adj[1].push_back(3);
    adj[2].push_back(4);
    adj[2].push_back(5);
    adj[2].push_back(6);
    adj[3].push_back(7);
    adj[5].push_back(8);
    adj[5].push_back(9);
    adj[5].push_back(10);
 
    unordered_set<int> values;
 
    int ans = 0;
 
    dfs(val, n, adj, 1, values, ans);
 
    cout << ans;
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
static int ans;
 
// DFS function to traverse the tree and find
// number of special nodes
static void dfs(int val[], int n, Vector<Integer> adj[], int v,
        HashSet<Integer> values)
{
 
    // If value of current node is already
    // present earlier in path then this
    // node and all other nodes connected to
    // it are not special
    if (values.contains(val[v]))
        return;
 
    // Insert value of current node in
    // set of values traversed
    ans++;
    values.add(val[v]);
 
    // Call dfs on all adjacent nodes
    for (int ele : adj[v])
    {
        dfs(val, n, adj, ele, values);
    }
 
    // Erase value of current node as all paths
    // passing through current node are traversed
    values.remove(val[v]);
}
 
// Driver code
public static void main(String[] args)
{
    int val[] = { 0, 2, 1, 4, 3, 4, 8, 10, 2, 5, 1 };
    int n = val.length;
 
    Vector<Integer> []adj = new Vector[n];
    for(int i = 0; i < n ; i++)
    {
        adj[i] = new Vector<Integer>();
    }
    adj[1].add(2);
    adj[1].add(3);
    adj[2].add(4);
    adj[2].add(5);
    adj[2].add(6);
    adj[3].add(7);
    adj[5].add(8);
    adj[5].add(9);
    adj[5].add(10);
 
    ans = 0;
    dfs(val, n, adj, 1, new HashSet<Integer>());
    System.out.print(ans);
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# DFS function to traverse the tree
# and find number of special nodes
def dfs(val, n, adj, v, values):
 
    # If value of current node is already
    # present earlier in path then this
    # node and all other nodes connected
    # to it are not special
    if val[v] in values:
        return
     
    global ans
 
    # Insert value of current node in
    # set of values traversed
    ans += 1
    values.add(val[v])
 
    # Call dfs on all adjacent nodes
    for ele in adj[v]:
        dfs(val, n, adj, ele, values)
 
    # Erase value of current node as all
    # paths passing through current node
    # are traversed
    values.remove(val[v])
 
# Driver code
if __name__ == "__main__":
 
    val = [0, 2, 1, 4, 3, 4, 8, 10, 2, 5, 1]
    n = len(val)
 
    adj = [[] for i in range(n)]
 
    adj[1].append(2)
    adj[1].append(3)
    adj[2].append(4)
    adj[2].append(5)
    adj[2].append(6)
    adj[3].append(7)
    adj[5].append(8)
    adj[5].append(9)
    adj[5].append(10)
 
    values = set()
    ans = 0
    dfs(val, n, adj, 1, values)
    print(ans)
 
# This code is contributed by Rituraj Jain

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
  
static int ans;
  
// DFS function to traverse the tree and find
// number of special nodes
static void dfs(int []val, int n, List<int> []adj, int v,
        HashSet<int> values)
{
  
    // If value of current node is already
    // present earlier in path then this
    // node and all other nodes connected to
    // it are not special
    if (values.Contains(val[v]))
        return;
  
    // Insert value of current node in
    // set of values traversed
    ans++;
    values.Add(val[v]);
  
    // Call dfs on all adjacent nodes
    foreach (int ele in adj[v])
    {
        dfs(val, n, adj, ele, values);
    }
  
    // Erase value of current node as all paths
    // passing through current node are traversed
    values.Remove(val[v]);
}
  
// Driver code
public static void Main(String[] args)
{
    int []val = { 0, 2, 1, 4, 3, 4, 8, 10, 2, 5, 1 };
    int n = val.Length;
  
    List<int> []adj = new List<int>[n];
    for(int i = 0; i < n ; i++)
    {
        adj[i] = new List<int>();
    }
    adj[1].Add(2);
    adj[1].Add(3);
    adj[2].Add(4);
    adj[2].Add(5);
    adj[2].Add(6);
    adj[3].Add(7);
    adj[5].Add(8);
    adj[5].Add(9);
    adj[5].Add(10);
  
    ans = 0;
    dfs(val, n, adj, 1, new HashSet<int>());
    Console.Write(ans);
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// JavaScript implementation of the approach
 
    var ans;
 
    // DFS function to traverse the tree and find
    // number of special nodes
    function dfs(val , n, adj , v,  values) {
 
        // If value of current node is already
        // present earlier in path then this
        // node and all other nodes connected to
        // it are not special
        if (values.has(val[v]))
            return;
 
        // Insert value of current node in
        // set of values traversed
        ans++;
        values.add(val[v]);
 
        // Call dfs on all adjacent nodes
        adj[v].forEach(function(ele) {
    
            dfs(val, n, adj, ele, values);
        });
 
        // Erase value of current node as all paths
        // passing through current node are traversed
        values.delete(val[v]);
    }
 
    // Driver code
     
        var val = [ 0, 2, 1, 4, 3, 4, 8, 10, 2, 5, 1 ];
        var n = val.length;
 
        var adj = [];
        for (var i = 0; i < n; i++) {
            adj[i] = [];
        }
        adj[1].push(2);
        adj[1].push(3);
        adj[2].push(4);
        adj[2].push(5);
        adj[2].push(6);
        adj[3].push(7);
        adj[5].push(8);
        adj[5].push(9);
        adj[5].push(10);
 
        ans = 0;
        dfs(val, n, adj, 1, new Set());
        document.write(ans);
 
// This code contributed by aashish1995
 
</script>
Output: 
8

 

Time Complexity: O(n)
 

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