Given a parent array P, where P[i] indicates the parent of ith node in the tree(assume parent of root node id indicated with -1). Find the height of the tree.
Examples:
Input : array[] = [-1 0 1 6 6 0 0 2 7]
Output : height = 5
Tree formed is:
0
/ | \
5 1 6
/ | \
2 4 3
/
7
/
8
1. Start at each node and keep going to its parent until we reach -1.
2. Also keep track of the maximum height among all nodes.
C++
// C++ program to find the height of the generic // tree(n-ary tree) if parent array is given #include <bits/stdc++.h> using namespace std; // function to find the height of tree int findHeight(int* parent, int n) { int res = 0; // Traverse each node for (int i = 0; i < n; i++) { // traverse to parent until -1 // is reached int p = i, current = 1; while (parent[p] != -1) { current++; p = parent[p]; } res = max(res, current); } return res; } // Driver program int main() { int parent[] = { -1, 0, 1, 6, 6, 0, 0, 2, 7 }; int n = sizeof(parent) / sizeof(parent[0]); int height = findHeight(parent, n); cout << "Height of the given tree is: " << height << endl; return 0; } |
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Java
// Java program to find the height of // the generic tree(n-ary tree) if // parent array is given import java.io.*; public class GFG { // function to find the height of tree static int findHeight(int[] parent, int n) { int res = 0; // Traverse each node for (int i = 0; i < n; i++) { // traverse to parent until -1 // is reached int p = i, current = 1; while (parent[p] != -1) { current++; p = parent[p]; } res = Math.max(res, current); } return res; } // Driver program static public void main(String[] args) { int[] parent = { -1, 0, 1, 6, 6, 0, 0, 2, 7 }; int n = parent.length; int height = findHeight(parent, n); System.out.println("Height of the " + "given tree is: " + height); } } // This code is contributed by vt_m. |
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Python3
# Python program to find the height of the generic # tree(n-ary tree) if parent array is given # function to find the height of tree def findHeight(parent, n): res = 0 # Traverse each node for i in range(n): # traverse to parent until -1 # is reached p = i current = 1 while (parent[p] != -1): current+= 1 p = parent[p] res = max(res, current) return res # Driver code if __name__ == '__main__': parent = [-1, 0, 1, 6, 6, 0, 0, 2, 7] n = len(parent) height = findHeight(parent, n) print("Height of the given tree is:", height) # This code is contributed by SHUBHAMSINGH10 |
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C#
// C# program to find the height of // the generic tree(n-ary tree) if // parent array is given using System; public class GFG { // function to find the height of tree static int findHeight(int[] parent, int n) { int res = 0; // Traverse each node for (int i = 0; i < n; i++) { // traverse to parent until -1 // is reached int p = i, current = 1; while (parent[p] != -1) { current++; p = parent[p]; } res = Math.Max(res, current); } return res; } // Driver program static public void Main() { int[] parent = { -1, 0, 1, 6, 6, 0, 0, 2, 7 }; int n = parent.Length; int height = findHeight(parent, n); Console.WriteLine("Height of the " + "given tree is: " + height); } } // This code is contributed by vt_m. |
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Output:
Height of the given tree is: 5
Optimized approach
We use dynamic programming. We store height from root to each node in an array.
So if we know height of root to a node then we can get height from root to nodes child by simply adding 1.
// C++ program to find the height of the generic // tree(n-ary tree) if parent array is given #include <bits/stdc++.h> using namespace std; // function to fill the height vector int rec(int i, int parent[], vector<int> height) { // if we have reached root node the // return 1 as height of root node if (parent[i] == -1) { return 1; } // if we have calculated height of a // node then return if if (height[i] != -1) { return height[i]; } // height from root to a node = height // from root to nodes parent + 1 height[i] = rec(parent[i], parent, height) + 1; // return nodes height return height[i]; } // function to find the height of tree int findHeight(int* parent, int n) { int res = 0; // vector to store heights of all nodes vector<int> height(n, -1); for (int i = 0; i < n; i++) { res = max(res, rec(i, parent, height)); } return res; } // Driver program int main() { int parent[] = { -1, 0, 1, 6, 6, 0, 0, 2, 7 }; int n = sizeof(parent) / sizeof(parent[0]); int height = findHeight(parent, n); cout << "Height of the given tree is: " << height << endl; return 0; } |
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Output:
Height of the given tree is: 5
Time complexity :- O(n)
Space complexity :- O(n)
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