Number of special nodes in an n-ary tree

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++

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// 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;
}

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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

Output:

8

Time Complexity: O(n)



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Improved By : rituraj_jain