Given an N-ary tree and an integer K, the task is to print the Kth ancestors of all the nodes of the tree in level order manner. If K ancestors does not exist for a node, then print -1 for that node.
Examples:
Input: K = 2
Output: -1 -1 -1 1 1 1 1 1 1
Explanation:
2nd ancestor does not exist for nodes 1, 2 and 3
2nd ancestor of nodes 4, 5, 6, 7, 8, 9 is 1.Input: K = 1
Output: -1 1 1 2 2 2 3 3 3
Approach: The approach is to use DFS to find the ancestors of all the nodes. Below are the steps:
- The Kth parent of any node can be found by using DFS, and storing all parents of a node in a temporary vector say temp[].
- Whenever a node is visited in DFS, it is pushed in the temp vector.
- At the end of DFS, the currently visited node is popped from the temp vector.
- For the currently visited node, the vector contains all the ancestors of the node.
- The Kth node from the end of the vector is the Kth ancestor of the currently visited node, so store it in a ancestor[] array.
Below is the implementation of the above approach:
C++
// C++ implementation of// the above approach#include <bits/stdc++.h>using namespace std;// Function to add an// edge in the treevoid addEdge(vector<int> v[], int x, int y){ v[x].push_back(y); v[y].push_back(x);}// DFS to find the Kth// ancestor of every nodevoid dfs(vector<int> tree[], vector<int>& temp, int ancestor[], int u, int parent, int k){ // Pushing current node // in the vector temp.push_back(u); // Traverse its neighbors for (auto i : tree[u]) { if (i == parent) continue; dfs(tree, temp, ancestor, i, u, k); } temp.pop_back(); // If K ancestors are not // found for current node if (temp.size() < k) { ancestor[u] = -1; } else { // Add the Kth ancestor // for the node ancestor[u] = temp[temp.size() - k]; }}// Function to find Kth// ancestor of each nodevoid KthAncestor(int N, int K, int E, int edges[][2]){ // Building the tree vector<int> tree[N + 1]; for (int i = 0; i < E; i++) { addEdge(tree, edges[i][0], edges[i][1]); } // Stores all parents of a node vector<int> temp; // Store Kth ancestor // of all nodes int ancestor[N + 1]; dfs(tree, temp, ancestor, 1, 0, K); // Print the ancestors for (int i = 1; i <= N; i++) { cout << ancestor[i] << " "; }}int main(){ // Given N and K int N = 9; int K = 2; // Given edges of n-ary tree int E = 8; int edges[8][2] = { { 1, 2 }, { 1, 3 }, { 2, 4 }, { 2, 5 }, { 2, 6 }, { 3, 7 }, { 3, 8 }, { 3, 9 } }; // Function Call KthAncestor(N, K, E, edges); return 0;} |
Java
// Java implementation of// the above approachimport java.util.*;class GFG{// Function to add an// edge in the treestatic void addEdge(Vector<Integer> v[], int x, int y){ v[x].add(y); v[y].add(x);}// DFS to find the Kth// ancestor of every nodestatic void dfs(Vector<Integer> tree[], Vector<Integer> temp, int ancestor[], int u, int parent, int k){ // Pushing current node // in the vector temp.add(u); // Traverse its neighbors for(int i : tree[u]) { if (i == parent) continue; dfs(tree, temp, ancestor, i, u, k); } temp.remove(temp.size() - 1); // If K ancestors are not // found for current node if (temp.size() < k) { ancestor[u] = -1; } else { // Add the Kth ancestor // for the node ancestor[u] = temp.get(temp.size() - k); }}// Function to find Kth// ancestor of each nodestatic void KthAncestor(int N, int K, int E, int edges[][]){ // Building the tree @SuppressWarnings("unchecked") Vector<Integer> []tree = new Vector[N + 1]; for(int i = 0; i < tree.length; i++) tree[i] = new Vector<Integer>(); for(int i = 0; i < E; i++) { addEdge(tree, edges[i][0], edges[i][1]); } // Stores all parents of a node Vector<Integer> temp = new Vector<Integer>(); // Store Kth ancestor // of all nodes int []ancestor = new int[N + 1]; dfs(tree, temp, ancestor, 1, 0, K); // Print the ancestors for(int i = 1; i <= N; i++) { System.out.print(ancestor[i] + " "); }}// Driver codepublic static void main(String[] args){ // Given N and K int N = 9; int K = 2; // Given edges of n-ary tree int E = 8; int edges[][] = { { 1, 2 }, { 1, 3 }, { 2, 4 }, { 2, 5 }, { 2, 6 }, { 3, 7 }, { 3, 8 }, { 3, 9 } }; // Function call KthAncestor(N, K, E, edges);}}// This code is contributed by Amit Katiyar |
Python3
# Python3 implementation of# the above approach# Function to add an# edge in the treedef addEdge(v, x, y): v[x].append(y) v[y].append(x)# DFS to find the Kth# ancestor of every nodedef dfs(tree, temp, ancestor, u, parent, k): # Pushing current node # in the vector temp.append(u) # Traverse its neighbors for i in tree[u]: if (i == parent): continue dfs(tree, temp, ancestor, i, u, k) temp.pop() # If K ancestors are not # found for current node if (len(temp) < k): ancestor[u] = -1 else: # Add the Kth ancestor # for the node ancestor[u] = temp[len(temp) - k]# Function to find Kth# ancestor of each nodedef KthAncestor(N, K, E, edges): # Building the tree tree = [[] for i in range(N + 1)] for i in range(E): addEdge(tree, edges[i][0], edges[i][1]) # Stores all parents of a node temp = [] # Store Kth ancestor # of all nodes ancestor = [0] * (N + 1) dfs(tree, temp, ancestor, 1, 0, K) # Print the ancestors for i in range(1, N + 1): print(ancestor[i], end = " ")# Driver codeif __name__ == '__main__': # Given N and K N = 9 K = 2 # Given edges of n-ary tree E = 8 edges = [ [ 1, 2 ], [ 1, 3 ], [ 2, 4 ], [ 2, 5 ], [ 2, 6 ], [ 3, 7 ], [ 3, 8 ], [ 3, 9 ] ] # Function call KthAncestor(N, K, E, edges)# This code is contributed by mohit kumar 29 |
C#
// C# implementation of// the above approachusing System;using System.Collections.Generic;class GFG{// Function to add an// edge in the treestatic void addEdge(List<int> []v, int x, int y){ v[x].Add(y); v[y].Add(x);}// DFS to find the Kth// ancestor of every nodestatic void dfs(List<int> []tree, List<int> temp, int []ancestor, int u, int parent, int k){ // Pushing current node // in the vector temp.Add(u); // Traverse its neighbors foreach(int i in tree[u]) { if (i == parent) continue; dfs(tree, temp, ancestor, i, u, k); } temp.RemoveAt(temp.Count - 1); // If K ancestors are not // found for current node if (temp.Count < k) { ancestor[u] = -1; } else { // Add the Kth ancestor // for the node ancestor[u] = temp[temp.Count - k]; }}// Function to find Kth// ancestor of each nodestatic void KthAncestor(int N, int K, int E, int [,]edges){ // Building the tree List<int> []tree = new List<int>[N + 1]; for(int i = 0; i < tree.Length; i++) tree[i] = new List<int>(); for(int i = 0; i < E; i++) { addEdge(tree, edges[i, 0], edges[i, 1]); } // Stores all parents of a node List<int> temp = new List<int>(); // Store Kth ancestor // of all nodes int []ancestor = new int[N + 1]; dfs(tree, temp, ancestor, 1, 0, K); // Print the ancestors for(int i = 1; i <= N; i++) { Console.Write(ancestor[i] + " "); }}// Driver codepublic static void Main(String[] args){ // Given N and K int N = 9; int K = 2; // Given edges of n-ary tree int E = 8; int [,]edges = { { 1, 2 }, { 1, 3 }, { 2, 4 }, { 2, 5 }, { 2, 6 }, { 3, 7 }, { 3, 8 }, { 3, 9 } }; // Function call KthAncestor(N, K, E, edges);}}// This code is contributed by Amit Katiyar |
Javascript
<script> // JavaScript program for the above approach // Function to add an // edge in the tree function addEdge(v, x, y) { v[x].push(y); v[y].push(x); } // DFS to find the Kth // ancestor of every node function dfs(tree, temp, ancestor, u, parent, k) { // Pushing current node // in the vector temp.push(u); // Traverse its neighbors for(let i = 0; i < tree[u].length; i++) { if (tree[u][i] == parent) continue; dfs(tree, temp, ancestor, tree[u][i], u, k); } temp.pop(); // If K ancestors are not // found for current node if (temp.length < k) { ancestor[u] = -1; } else { // Add the Kth ancestor // for the node ancestor[u] = temp[temp.length - k]; } } // Function to find Kth // ancestor of each node function KthAncestor(N, K, E, edges) { // Building the tree let tree = new Array(N + 1); for(let i = 0; i < tree.length; i++) tree[i] = []; for(let i = 0; i < E; i++) { addEdge(tree, edges[i][0], edges[i][1]); } // Stores all parents of a node let temp = []; // Store Kth ancestor // of all nodes let ancestor = new Array(N + 1); dfs(tree, temp, ancestor, 1, 0, K); // Print the ancestors for(let i = 1; i <= N; i++) { document.write(ancestor[i] + " "); } } // Given N and K let N = 9; let K = 2; // Given edges of n-ary tree let E = 8; let edges = [ [ 1, 2 ], [ 1, 3 ], [ 2, 4 ], [ 2, 5 ], [ 2, 6 ], [ 3, 7 ], [ 3, 8 ], [ 3, 9 ] ]; // Function call KthAncestor(N, K, E, edges); </script> |
Output:
-1 -1 -1 1 1 1 1 1 1
Time complexity: O(N)
Auxiliary Space: O(N)
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