打印从节点到给定完全二叉树的根的路径

给定一个整数N ，任务是找到从第N^个节点到以下形式的二叉树根的路径：

• The Binary Tree is a Complete Binary Tree up to the level of the N^th node.
• The nodes are numbered 1 to N, starting from the root as 1.
• The structure of the Tree is as follows: 
   

例子：

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.

朴素的方法：解决问题的最简单的方法是执行DFS 从给定节点直到遇到根节点并打印路径。

时间复杂度： O(N)
辅助空间： O(1)

高效的方法：可以根据给定的二叉树的结构优化上述方法。可以观察到，对于每个N ，其父节点将是N / 2 。因此，反复打印的N和更新N到N / 2的电流值，直到N等于1时，达到即根节点。

下面是上述方法的实现：

C++

1
           /       \
          2         3
       /    \    /   \
      4     5    6    7
      ................
   /    \ ............
 N - 1  N ............

Java

// 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;
}

Python3

// 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

C#

# 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

Javascript

// 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

输出：

 
// 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
 

时间复杂度： O(log ₂ (N))
辅助空间： O(1)

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