Given an integer N, the task is to find the path from the Nth node to the root of a Binary Tree of the following form:
- The Binary Tree is a Complete Binary Tree up to the level of the Nth node.
- The nodes are numbered 1 to N, starting from the root as 1.
- The structure of the Tree is as follows:
1 / \ 2 3 / \ / \ 4 5 6 7 ................ / \ ............ N - 1 N ............
Examples:
Input: N = 7
Output: 7 3 1
Explanation: The path from the node 7 to root is 7 -> 3 -> 1.Input: N = 11
Output: 11 5 2 1
Explanation: The path from node 11 to root is 11 -> 5 -> 2 -> 1.
Naive Approach: The simplest approach to solve the problem is to perform DFS from the given node until the root node is encountered and print the path.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the structure of the given Binary Tree. It can be observed that for every N, its parent node will be N / 2. Therefore, repeatedly print the current value of N and update N to N / 2 until N is equal to 1, i.e. root node is reached.
Below is the implementation of the above approach:
// C++ program for the above approach #include <iostream> using namespace std;
// Function to print the path // from node to root void path_to_root( int node)
{ // Iterate until root is reached
while (node >= 1) {
// Print the value of
// the current node
cout << node << ' ' ;
// Move to parent of
// the current node
node /= 2;
}
} // Driver Code int main()
{ int N = 7;
path_to_root(N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to print the path // from node to root static void path_to_root( int node)
{ // Iterate until root is reached
while (node >= 1 )
{
// Print the value of
// the current node
System.out.print(node + " " );
// Move to parent of
// the current node
node /= 2 ;
}
} // Driver Code public static void main(String[] args)
{ int N = 7 ;
path_to_root(N);
} } // This code is contributed by shivanisinghss2110 |
# Python3 program for the above approach # Function to print the path # from node to root def path_to_root(node):
# Iterate until root is reached
while (node > = 1 ):
# Print the value of
# the current node
print (node, end = " " )
# Move to parent of
# the current node
node / / = 2
# Driver Code if __name__ = = '__main__' :
N = 7
path_to_root(N)
# This code is contributed by mohit kumar 29 |
// C# program for the above approach using System;
class GFG
{ // Function to print the path // from node to root static void path_to_root( int node)
{ // Iterate until root is reached
while (node >= 1)
{
// Print the value of
// the current node
Console.Write(node + " " );
// Move to parent of
// the current node
node /= 2;
}
} // Driver Code public static void Main(String[] args)
{ int N = 7;
path_to_root(N);
} } // This code is contributed by shivanisinghss2110 |
<script> // Javascript program for the above approach // Function to print the path // from node to root function path_to_root(node)
{ // Iterate until root is reached
while (node >= 1)
{
// Print the value of
// the current node
document.write(node + " " );
// Move to parent of
// the current node
node = parseInt(node / 2, 10);
}
} // Driver code let N = 7; path_to_root(N); // This code is contributed by divyeshrabadiya07 </script> |
7 3 1
Time Complexity: O(log2(N))
Auxiliary Space: O(1)