给定一个数字N ，它是一个完整的二叉树中节点的总数，其中节点的级别是1到N。任务是编写一个程序，以打印从根到“完整二进制树”中所有节点的路径。
对于N = 3，树将是：
1
/ \
2 3
对于N = 7，树将为：
1
/ \
2 3
/ \ / \
4 5 6 7
例子：
Input : 7
Output :
1
1 2
1 2 4
1 2 5
1 3
1 3 6
1 3 7
Input : 4
Output :
1
1 2
1 2 4
1 3
说明：-因为，给定的树是完整的二叉树。对于每个节点
我们可以计算出它的左孩子为2 * i ，右孩子为2 * i +1 。
这个想法是使用回溯方法来打印所有路径。维护一个向量来存储路径，首先将根节点1推入其中，然后在推动左子节点和右子节点之前，先打印存储在其中的当前路径，然后再为左子节点和右子节点调用该函数。
以下是上述方法的完整实现：
C++
// C++ program to print path from root to all
// nodes in a complete binary tree.
#include
#include
using namespace std;
// Function to print path of all the nodes
// nth node represent as given node
// kth node represents as left and right node
void printPath(vector res, int nThNode, int kThNode)
{
// base condition
// if kth node value is greater
// then nth node then its means
// kth node is not valid so
// we not store it into the res
// simply we just return
if (kThNode > nThNode)
return;
// Storing node into res
res.push_back(kThNode);
// Print the path from root to node
for (int i = 0; i < res.size(); i++)
cout << res[i] << " ";
cout << "\n";
// store left path of a tree
// So for left we will go node(kThNode*2)
printPath(res, nThNode, kThNode * 2);
// right path of a tree
// and for right we will go node(kThNode*2+1)
printPath(res, nThNode, kThNode * 2 + 1);
}
// Function to print path from root to all of the nodes
void printPathToCoverAllNodeUtil(int nThNode)
{
// res is for store the path
// from root to particulate node
vector res;
// Print path from root to all node.
// third argument 1 because of we have
// to consider root node is 1
printPath(res, nThNode, 1);
}
// Driver Code
int main()
{
// Given Node
int nThNode = 7;
// Print path from root to all node.
printPathToCoverAllNodeUtil(nThNode);
return 0;
} Java
// Java program to print path from root to all
// nodes in a complete binary tree.
import java.util.*;
class GFG
{
// Function to print path of all the nodes
// nth node represent as given node
// kth node represents as left and right node
static void printPath(Vector res,
int nThNode, int kThNode)
{
// base condition
// if kth node value is greater
// then nth node then its means
// kth node is not valid so
// we not store it into the res
// simply we just return
if (kThNode > nThNode)
return;
// Storing node into res
res.add(kThNode);
// Print the path from root to node
for (int i = 0; i < res.size(); i++)
System.out.print( res.get(i) + " ");
System.out.print( "\n");
// store left path of a tree
// So for left we will go node(kThNode*2)
printPath(res, nThNode, kThNode * 2);
// right path of a tree
// and for right we will go node(kThNode*2+1)
printPath(res, nThNode, kThNode * 2 + 1);
res.remove(res.size()-1);
}
// Function to print path from root to all of the nodes
static void printPathToCoverAllNodeUtil(int nThNode)
{
// res is for store the path
// from root to particulate node
Vector res=new Vector();
// Print path from root to all node.
// third argument 1 because of we have
// to consider root node is 1
printPath(res, nThNode, 1);
}
// Driver Code
public static void main(String args[])
{
// Given Node
int nThNode = 7;
// Print path from root to all node.
printPathToCoverAllNodeUtil(nThNode);
}
}
// This code is contributed by Arnab Kundu Python3
# Python3 program to print path from root
# to all nodes in a complete binary tree.
# Function to print path of all the nodes
# nth node represent as given node kth
# node represents as left and right node
def printPath(res, nThNode, kThNode):
# base condition
# if kth node value is greater
# then nth node then its means
# kth node is not valid so
# we not store it into the res
# simply we just return
if kThNode > nThNode:
return
# Storing node into res
res.append(kThNode)
# Print the path from root to node
for i in range(0, len(res)):
print(res[i], end = " ")
print()
# store left path of a tree
# So for left we will go node(kThNode*2)
printPath(res[:], nThNode, kThNode * 2)
# right path of a tree
# and for right we will go node(kThNode*2+1)
printPath(res[:], nThNode, kThNode * 2 + 1)
# Function to print path from root
# to all of the nodes
def printPathToCoverAllNodeUtil(nThNode):
# res is for store the path
# from root to particulate node
res = []
# Print path from root to all node.
# third argument 1 because of we have
# to consider root node is 1
printPath(res, nThNode, 1)
# Driver Code
if __name__ == "__main__":
# Given Node
nThNode = 7
# Print path from root to all node.
printPathToCoverAllNodeUtil(nThNode)
# This code is contributed by Rituraj JainC#
// C# program to print path from root to all
// nodes in a complete binary tree.
using System;
using System.Collections.Generic;
class GFG
{
// Function to print path of all the nodes
// nth node represent as given node
// kth node represents as left and right node
static void printPath(List res,
int nThNode, int kThNode)
{
// base condition
// if kth node value is greater
// then nth node then its means
// kth node is not valid so
// we not store it into the res
// simply we just return
if (kThNode > nThNode)
return;
// Storing node into res
res.Add(kThNode);
// Print the path from root to node
for (int i = 0; i < res.Count; i++)
Console.Write( res[i] + " ");
Console.Write( "\n");
// store left path of a tree
// So for left we will go node(kThNode*2)
printPath(res, nThNode, kThNode * 2);
// right path of a tree
// and for right we will go node(kThNode*2+1)
printPath(res, nThNode, kThNode * 2 + 1);
res.RemoveAt(res.Count-1);
}
// Function to print path from root to all of the nodes
static void printPathToCoverAllNodeUtil(int nThNode)
{
// res is for store the path
// from root to particulate node
List res=new List();
// Print path from root to all node.
// third argument 1 because of we have
// to consider root node is 1
printPath(res, nThNode, 1);
}
// Driver Code
public static void Main(String []args)
{
// Given Node
int nThNode = 7;
// Print path from root to all node.
printPathToCoverAllNodeUtil(nThNode);
}
}
// This code contributed by Rajput-Ji PHP
$nThNode)
return;
// Storing node into res
array_push($res, $kThNode);
// Print the path from root to node
for ($i = 0; $i < count($res); $i++)
echo $res[$i] . " ";
echo "\n";
// store left path of a tree
// So for left we will go node(kThNode*2)
printPath($res, $nThNode, $kThNode * 2);
// right path of a tree
// and for right we will go node(kThNode*2+1)
printPath($res, $nThNode, $kThNode * 2 + 1);
}
// Function to print path
// from root to all of the nodes
function printPathToCoverAllNodeUtil($nThNode)
{
// res is for store the path
// from root to particulate node
$res = array();
// Print path from root to all node.
// third argument 1 because of we have
// to consider root node is 1
printPath($res, $nThNode, 1);
}
// Driver Code
// Given Node
$nThNode = 7;
// Print path from root to all node.
printPathToCoverAllNodeUtil($nThNode);
// This code is contributed by mits
?>输出：
1
1 2
1 2 4
1 2 5
1 3
1 3 6
1 3 7
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