Count the number of common ancestors of given K nodes in a N-ary Tree

Given an N-ary tree root and a list of K nodes, the task is to find the number of common ancestors of the given K nodes in the tree.

Example:

Input: root = 3  
                    /   \   
                 2      1
               / \    / | \   
             9  7  8  6 3
K = {7, 2, 9}
Output: 2
Explanation: The common ancestors of the nodes 7, 9 and 2 are 2 and 3

Input: root = 2 
                       \
                        1
                         \
                          0—4
                        / | \
                      9  3 8
K = {9, 8, 3, 4, 0}
Output: 3

Approach: The given problem can be solved by using the post-order traversal. The idea is to find the lowest common ancestor of the K nodes then increment the count of ancestors for every node above it till the root is reached. Below steps can be followed to solve the problem:

• Add all the list nodes into a set
• Apply post-order traversal on the tree:
  • Find the lowest common ancestor then start incrementing the value of the number of nodes at every recursive call
• Return the answer calculated

C++

// C++ implementation for the above approach
#include<bits/stdc++.h>
using namespace std;

class Node {

  public:
  vector<Node*> children;
  int val;
  Node(int val)
  {
    this->val = val;
  }
};

// Function to find LCA and
// count number of ancestors
vector<Node*> CAcount(Node* root, set<Node*> st)
{

  // If the current node
  // is a desired node
  if (st.count(root)) {

    vector<Node*> res(2, NULL);
    res[0] = root;
    res[1] = new Node(1);
    return res;
  }

  // If leaf node then return null
  if (root->children.size() == 0) {

    vector<Node*> res(2, NULL);
    return res;
  }

  // To count number of desired nodes
  // in the children branches
  int childCount = 0;

  // Initialize a node to return
  vector<Node*> ans(2, NULL);


  // Iterate through all children
  for(auto child: root->children){

    vector<Node*> res = CAcount(child, st);

    // Increment child count if
    // desired node is found
    if (res[0] != NULL)
      childCount++;

    // If first desired node is found
    if (childCount == 1 && ans[0] == NULL) {

      ans = res;
    }
    else if (childCount > 1) {

      ans[0] = root;
      ans[1] = new Node(1);
      return ans;
    }
  }

  // If LCA found below then increment
  // number of common ancestors
  if (ans[0] != NULL)
    ans[1]->val++;

  // Return the answer
  return ans;
}


// Function to find the number
// of common ancestors in a tree
int numberOfAncestors(Node* root, vector<Node*> nodes)
{

  // Initialize a set
  set<Node*> st;


  // Iterate the list of nodes
  // and add them in a set

  for (auto curr: nodes){
    st.insert(curr);
  }

  // Find LCA and return
  // number of ancestors
  return CAcount(root, st)[1]->val;
}



// Driver code
int main()
{

  // Initialize the tree
  Node* zero = new Node(0);
  Node* one = new Node(1);
  Node* two = new Node(2);
  Node* three = new Node(3);
  Node* four = new Node(4);
  Node* five = new Node(5);
  Node* six = new Node(6);
  Node* seven = new Node(7);
  zero->children.push_back(one);
  zero->children.push_back(two);
  zero->children.push_back(three);
  one->children.push_back(four);
  one->children.push_back(five);
  five->children.push_back(six);
  five->children.push_back(seven);


  // List of nodes whose
  // ancestors are to be found
  vector<Node*> nodes;
  nodes.push_back(four);
  nodes.push_back(six);
  nodes.push_back(seven);

  // Call the function
  // and print the result
  cout << numberOfAncestors(zero, nodes) << endl;

  return 0;
}

// The code is contributed by Nidhi goel.

Java

// Java implementation for the above approach

import java.io.*;
import java.util.*;

class GFG {

    static class Node {

        List<Node> children;
        int val;

        // constructor
        public Node(int val)
        {
            children = new ArrayList<>();
            this.val = val;
        }
    }

    // Function to find the number
    // of common ancestors in a tree
    public static int numberOfAncestors(
        Node root,
        List<Node> nodes)
    {

        // Initialize a set
        Set<Node> set = new HashSet<>();

        // Iterate the list of nodes
        // and add them in a set
        for (Node curr : nodes) {
            set.add(curr);
        }

        // Find LCA and return
        // number of ancestors
        return CAcount(root, set)[1].val;
    }

    // Function to find LCA and
    // count number of ancestors
    public static Node[] CAcount(
        Node root, Set<Node> set)
    {

        // If the current node
        // is a desired node
        if (set.contains(root)) {

            Node[] res = new Node[2];
            res[0] = root;
            res[1] = new Node(1);
            return res;
        }

        // If leaf node then return null
        if (root.children.size() == 0) {

            return new Node[2];
        }

        // To count number of desired nodes
        // in the children branches
        int childCount = 0;

        // Initialize a node to return
        Node[] ans = new Node[2];

        // Iterate through all children
        for (Node child : root.children) {

            Node[] res = CAcount(child, set);

            // Increment child count if
            // desired node is found
            if (res[0] != null)
                childCount++;

            // If first desired node is found
            if (childCount == 1
                && ans[0] == null) {

                ans = res;
            }
            else if (childCount > 1) {

                ans[0] = root;
                ans[1] = new Node(1);
                return ans;
            }
        }

        // If LCA found below then increment
        // number of common ancestors
        if (ans[0] != null)
            ans[1].val++;

        // Return the answer
        return ans;
    }

    // Driver code
    public static void main(String[] args)
    {

        // Initialize the tree
        Node zero = new Node(0);
        Node one = new Node(1);
        Node two = new Node(2);
        Node three = new Node(3);
        Node four = new Node(4);
        Node five = new Node(5);
        Node six = new Node(6);
        Node seven = new Node(7);
        zero.children.add(one);
        zero.children.add(two);
        zero.children.add(three);
        one.children.add(four);
        one.children.add(five);
        five.children.add(six);
        five.children.add(seven);

        // List of nodes whose
        // ancestors are to be found
        List<Node> nodes = new ArrayList<>();
        nodes.add(four);
        nodes.add(six);
        nodes.add(seven);

        // Call the function
        // and print the result
        System.out.println(
            numberOfAncestors(zero, nodes));
    }
}

Python3

# Python3 implementation for the above approach
class Node:
    def __init__(self, val):
        self.children = []
        self.val = val

# Function to find the number
# of common ancestors in a tree
def number_of_ancestors(root, nodes):
    set_nodes = set(nodes)

  # Find LCA and return
  # number of ancestors
    return CAcount(root, set_nodes)[1].val

# Function to find LCA and
# count number of ancestors
def CAcount(root, set_nodes):
    if root in set_nodes:
        res = [root, Node(1)]
        return res
    if not root.children:
        return [None, None]

   # To count number of desired nodes
   # in the children branches
    child_count = 0
    ans = [None, None]
    for child in root.children:
        res = CAcount(child, set_nodes)

        # Increment child count if
        # desired node is found
        if res[0] is not None:
            child_count += 1
        if child_count == 1 and ans[0] is None:
            ans = res
        elif child_count > 1:
            ans = [root, Node(1)]
            return ans

    # If LCA found below then increment
    # number of common ancestors
    if ans[0] is not None:
        ans[1].val += 1
    return ans

# Initialize the tree
zero = Node(0)
one = Node(1)
two = Node(2)
three = Node(3)
four = Node(4)
five = Node(5)
six = Node(6)
seven = Node(7)
zero.children.append(one)
zero.children.append(two)
zero.children.append(three)
one.children.append(four)
one.children.append(five)
five.children.append(six)
five.children.append(seven)

nodes = [four, six, seven]
print(number_of_ancestors(zero, nodes))

# This code is contributed by Potta Lokesh

C#

// C# implementation for the above approach
using System;
using System.Collections.Generic;

public class GFG {

    class Node {

        public List<Node> children;
        public int val;

        // constructor
        public Node(int val)
        {
            children = new List<Node>();
            this.val = val;
        }
    }

    // Function to find the number
    // of common ancestors in a tree
    static int numberOfAncestors(
        Node root,
        List<Node> nodes)
    {

        // Initialize a set
        HashSet<Node> set = new HashSet<Node>();

        // Iterate the list of nodes
        // and add them in a set
        foreach (Node curr in nodes) {
            set.Add(curr);
        }

        // Find LCA and return
        // number of ancestors
        return CAcount(root, set)[1].val;
    }

    // Function to find LCA and
    // count number of ancestors
    static Node[] CAcount(
        Node root, HashSet<Node> set)
    {

        // If the current node
        // is a desired node
        if (set.Contains(root)) {

            Node[] res = new Node[2];
            res[0] = root;
            res[1] = new Node(1);
            return res;
        }

        // If leaf node then return null
        if (root.children.Count == 0) {

            return new Node[2];
        }

        // To count number of desired nodes
        // in the children branches
        int childCount = 0;

        // Initialize a node to return
        Node[] ans = new Node[2];

        // Iterate through all children
        foreach (Node child in root.children) {

            Node[] res = CAcount(child, set);

            // Increment child count if
            // desired node is found
            if (res[0] != null)
                childCount++;

            // If first desired node is found
            if (childCount == 1
                && ans[0] == null) {

                ans = res;
            }
            else if (childCount > 1) {

                ans[0] = root;
                ans[1] = new Node(1);
                return ans;
            }
        }

        // If LCA found below then increment
        // number of common ancestors
        if (ans[0] != null)
            ans[1].val++;

        // Return the answer
        return ans;
    }

    // Driver code
    public static void Main(String[] args)
    {

        // Initialize the tree
        Node zero = new Node(0);
        Node one = new Node(1);
        Node two = new Node(2);
        Node three = new Node(3);
        Node four = new Node(4);
        Node five = new Node(5);
        Node six = new Node(6);
        Node seven = new Node(7);
        zero.children.Add(one);
        zero.children.Add(two);
        zero.children.Add(three);
        one.children.Add(four);
        one.children.Add(five);
        five.children.Add(six);
        five.children.Add(seven);

        // List of nodes whose
        // ancestors are to be found
        List<Node> nodes = new List<Node>();
        nodes.Add(four);
        nodes.Add(six);
        nodes.Add(seven);

        // Call the function
        // and print the result
        Console.WriteLine(
            numberOfAncestors(zero, nodes));
    }
}

// This code is contributed by shikhasingrajput

Javascript

<script>
// Javascript implementation for the above approach



class Node {
  // constructor
  constructor(val) {
    this.children = new Array();
    this.val = val;
  }
}

// Function to find the number
// of common ancestors in a tree
function numberOfAncestors(root, nodes) {

  // Initialize a set
  let set = new Set();

  // Iterate the list of nodes
  // and add them in a set
  for (curr of nodes) {
    set.add(curr);
  }

  // Find LCA and return
  // number of ancestors
  return CAcount(root, set)[1].val;
}

// Function to find LCA and
// count number of ancestors
function CAcount(root, set) {

  // If the current node
  // is a desired node
  if (set.has(root)) {

    let res = new Node(2);
    res[0] = root;
    res[1] = new Node(1);
    return res;
  }

  // If leaf node then return null
  if (root.children.length == 0) {

    return new Node(2);
  }

  // To count number of desired nodes
  // in the children branches
  let childCount = 0;

  // Initialize a node to return
  let ans = new Node(2);

  // Iterate through all children
  for (child of root.children) {

    let res = CAcount(child, set);

    // Increment child count if
    // desired node is found
    if (res[0] != null)
      childCount++;

    // If first desired node is found
    if (childCount == 1
      && ans[0] == null) {

      ans = res;
    }
    else if (childCount > 1) {

      ans[0] = root;
      ans[1] = new Node(1);
      return ans;
    }
  }

  // If LCA found below then increment
  // number of common ancestors
  if (ans[0] != null)
    ans[1].val++;

  // Return the answer
  return ans;
}

// Driver code

// Initialize the tree
let zero = new Node(0);
let one = new Node(1);
let two = new Node(2);
let three = new Node(3);
let four = new Node(4);
let five = new Node(5);
let six = new Node(6);
let seven = new Node(7);
zero.children.push(one);
zero.children.push(two);
zero.children.push(three);
one.children.push(four);
one.children.push(five);
five.children.push(six);
five.children.push(seven);

// List of nodes whose
// ancestors are to be found
let nodes = new Array();
nodes.push(four);
nodes.push(six);
nodes.push(seven);

// Call the function
// and print the result
document.write(numberOfAncestors(zero, nodes));

// This code is contributed by saurabh_jaiswal.
</script>

Output

2

Time Complexity: O(N), where N is the number of nodes in the tree
Auxiliary Space: O(H), H is the height of the tree