给定一个 N -ary 树，任务是打印给定N-ary Tree 的所有根到叶路径。
例子：
Input:
1
/ \
2 3
/ / \
4 5 6
/ \
7 8
Output:
1 2 4
1 3 5
1 3 6 7
1 3 6 8
Input:
1
/ | \
2 5 3
/ \ \
4 5 6
Output:
1 2 4
1 2 5
1 5
1 3 6
方法：解决这个问题的想法是使用深度优先搜索开始遍历 N 叉树，并不断插入向量中遇到的每个节点，直到遇到叶节点。每当遇到叶节点时，将存储在向量中的元素作为当前遍历的根到叶路径打印并删除最后添加的叶并检查下一个组合。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Strcuture of an N ary tree node
class Node {
public:
int data;
vector child;
// Parameterized Constructor
Node(int x)
: data(x)
{
}
};
// Function to print the root to leaf
// path of the given N-ary Tree
void printPath(vector vec)
{
// Print elements in the vector
for (int ele : vec) {
cout << ele << " ";
}
cout << endl;
}
// Utility function to print all
// root to leaf paths of an Nary Tree
void printAllRootToLeafPaths(
Node* root, vector vec)
{
// If root is null
if (!root)
return;
// Insert current node's
// data into the vector
vec.push_back(root->data);
// If current node is a leaf node
if (root->child.empty()) {
// Print the path
printPath(vec);
// Pop the leaf node
// and return
vec.pop_back();
return;
}
// Recur for all children of
// the current node
for (int i = 0;
i < root->child.size(); i++)
// Recursive Function Call
printAllRootToLeafPaths(
root->child[i], vec);
}
// Function to print root to leaf path
void printAllRootToLeafPaths(Node* root)
{
// If root is null, return
if (!root)
return;
// Stores the root to leaf path
vector vec;
// Utility function call
printAllRootToLeafPaths(root, vec);
}
// Driver Code
int main()
{
// Given N-Ary tree
Node* root = new Node(1);
(root->child).push_back(new Node(2));
(root->child).push_back(new Node(3));
(root->child[0]->child).push_back(new Node(4));
(root->child[1]->child).push_back(new Node(5));
(root->child[1]->child).push_back(new Node(6));
(root->child[1]->child[1]->child)
.push_back(new Node(7));
(root->child[1]->child[1]->child)
.push_back(new Node(8));
// Function Call
printAllRootToLeafPaths(root);
return 0;
} Java
// Java program for the above approach
import java.util.ArrayList;
class GFG
{
// Strcuture of an N ary tree node
static class Node
{
int data;
ArrayList child;
// Parameterized Constructor
public Node(int x)
{
this.data = x;
this.child = new ArrayList<>();
}
};
// Function to print the root to leaf
// path of the given N-ary Tree
static void printPath(ArrayList vec)
{
// Print elements in the vector
for (int ele : vec)
{
System.out.print(ele + " ");
}
System.out.println();
}
// Utility function to print all
// root to leaf paths of an Nary Tree
static void printAllRootToLeafPaths(Node root, ArrayList vec)
{
// If root is null
if (root == null)
return;
// Insert current node's
// data into the vector
vec.add(root.data);
// If current node is a leaf node
if (root.child.isEmpty())
{
// Print the path
printPath(vec);
// Pop the leaf node
// and return
vec.remove(vec.size() - 1);
return;
}
// Recur for all children of
// the current node
for (int i = 0; i < root.child.size(); i++)
// Recursive Function Call
printAllRootToLeafPaths(root.child.get(i), vec);
vec.remove(vec.size() - 1);
}
// Function to print root to leaf path
static void printAllRootToLeafPaths(Node root)
{
// If root is null, return
if (root == null)
return;
// Stores the root to leaf path
ArrayList vec = new ArrayList<>();
// Utility function call
printAllRootToLeafPaths(root, vec);
}
// Driver Code
public static void main(String[] args)
{
// Given N-Ary tree
Node root = new Node(1);
(root.child).add(new Node(2));
(root.child).add(new Node(3));
(root.child.get(0).child).add(new Node(4));
(root.child.get(1).child).add(new Node(5));
(root.child.get(1).child).add(new Node(6));
(root.child.get(1).child.get(1).child).add(new Node(7));
(root.child.get(1).child.get(1).child).add(new Node(8));
// Function Call
printAllRootToLeafPaths(root);
}
}
// This code is contributed by sanjeev2552 Python3
# Python3 program for the above approach
# Strcuture of an N ary tree node
class Node:
def __init__(self, x):
self.data = x
self.child = []
# Function to print the root to leaf
# path of the given N-ary Tree
def printPath(vec):
# Print elements in the vector
for ele in vec:
print(ele, end = " ")
print()
# Utility function to print all
# root to leaf paths of an Nary Tree
def printAllRootToLeafPaths(root):
global vec
# If root is null
if (not root):
return
# Insert current node's
# data into the vector
vec.append(root.data)
# If current node is a leaf node
if (len(root.child) == 0):
# Print the path
printPath(vec)
# Pop the leaf node
# and return
vec.pop()
return
# Recur for all children of
# the current node
for i in range(len(root.child)):
# Recursive Function Call
printAllRootToLeafPaths(root.child[i])
vec.pop()
# Function to print root to leaf path
def printRootToLeafPaths(root):
global vec
# If root is null, return
if (not root):
return
# Utility function call
printAllRootToLeafPaths(root)
# Driver Code
if __name__ == '__main__':
# Given N-Ary tree
vec = []
root = Node(1)
root.child.append(Node(2))
root.child.append(Node(3))
root.child[0].child.append(Node(4))
root.child[1].child.append(Node(5))
root.child[1].child.append(Node(6))
root.child[1].child[1].child.append(Node(7))
root.child[1].child[1].child.append(Node(8))
# Function Call
printRootToLeafPaths(root)
# This code is contributed by mohit kumar 29C#
using System;
using System.Collections.Generic;
// Strcuture of an N ary tree node
public class Node
{
public int data;
public List child;
// Parameterized Constructor
public Node(int x)
{
this.data = x;
this.child = new List();
}
}
public class GFG
{
// Function to print the root to leaf
// path of the given N-ary Tree
static void printPath(List vec)
{
// Print elements in the vector
foreach (int ele in vec)
{
Console.Write(ele + " ");
}
Console.WriteLine();
}
// Utility function to print all
// root to leaf paths of an Nary Tree
static void printAllRootToLeafPaths(Node root, List vec)
{
// If root is null
if (root == null)
return;
// Insert current node's
// data into the vector
vec.Add(root.data);
// If current node is a leaf node
if (root.child.Count == 0)
{
// Print the path
printPath(vec);
// Pop the leaf node
// and return
vec.RemoveAt(vec.Count - 1);
return;
}
// Recur for all children of
// the current node
for (int i = 0; i < root.child.Count; i++)
{
// Recursive Function Call
printAllRootToLeafPaths(root.child[i], vec);
}
vec.RemoveAt(vec.Count - 1);
}
// Function to print root to leaf path
static void printAllRootToLeafPaths(Node root)
{
// If root is null, return
if (root == null)
return;
// Stores the root to leaf path
List vec = new List();
// Utility function call
printAllRootToLeafPaths(root, vec);
}
// Driver Code
static public void Main ()
{
// Given N-Ary tree
Node root = new Node(1);
(root.child).Add(new Node(2));
(root.child).Add(new Node(3));
(root.child[0].child).Add(new Node(4));
(root.child[1].child).Add(new Node(5));
(root.child[1].child).Add(new Node(6));
(root.child[1].child[1].child).Add(new Node(7));
(root.child[1].child[1].child).Add(new Node(8));
// Function Call
printAllRootToLeafPaths(root);
}
}
// This code is contributed by rag2127 输出：
1 2 4
1 3 5
1 3 6 7
1 3 6 8时间复杂度： O(N)
空间复杂度： O(N)
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