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 3Input: 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
Java
// Java implementation for the above approachimport 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)); }} |
C#
// C# implementation for the above approachusing 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 approachclass Node { // constructor constructor(val) { this.children = new Array(); this.val = val; }}// Function to find the number// of common ancestors in a treefunction 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 ancestorsfunction 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 treelet 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 foundlet nodes = new Array();nodes.push(four);nodes.push(six);nodes.push(seven);// Call the function// and print the resultdocument.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
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