原文:
给定一个 ，任务是。
示例:
输入: 1 /\ 2 3 /\ 4 5 6 /\ 7 8
产量:t2【1 2 4】t3【1 3 5】t4【1 3 6 7】t5【1 3 6 8】
输入: 1 /| \ 2 5 3 /\ \ 4 5 6 输出: 1 2 4 1 2 5 1 5 1 3 6
方法:解决这个问题的思路是使用开始，并保持 。每当遇到叶节点时，打印存储在中的元素，作为当前遍历的根到叶路径，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;
// structure 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 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.util.arraylist;
class gfg
{
// structure 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
python 3
# python3 program for the above approach
# structure 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 29
c
using system;
using system.collections.generic;
// structure 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
java 描述语言
output:
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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