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

• Difficulty Level : Expert
• Last Updated : 11 Jun, 2021

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 ``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

 ``
Output:
`7 3 1`

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

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