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Print path from a node to root of given Complete Binary Tree

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  • Difficulty Level : Medium
  • Last Updated : 11 Jun, 2021
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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++




// 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




// 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




# 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#




// 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

Javascript




<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>

Output: 

7 3 1

 

Time Complexity: O(log2(N))
Auxiliary Space: O(1)


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