# 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 3Output: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 1Output: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; ` `} ` |

*chevron_right*

*filter_none*

## 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)

## Recommended Posts:

- Find height of a special binary tree whose leaf nodes are connected
- Given a n-ary tree, count number of nodes which have more number of children than parents
- Number of nodes greater than a given value in n-ary tree
- Count the number of nodes at a given level in a tree using DFS
- Count the number of nodes at given level in a tree using BFS.
- Level with maximum number of nodes using DFS in a N-ary tree
- Reverse nodes of a linked list without affecting the special characters
- Number of leaf nodes in the subtree of every node of an n-ary tree
- Construct a special tree from given preorder traversal
- Print levels with odd number of nodes and even number of nodes
- Subtree of all nodes in a tree using DFS
- Product of all nodes in a Binary Tree
- Sum of all the Boundary Nodes of a Binary Tree
- Check if two nodes are on same path in a tree
- Convert a tree to forest of even nodes

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.