给定一个N元树，请打印每个节点的子树中的叶节点数。
例子：
Input:
1
/ \
2 3
/ | \
4 5 6
Output:
The node 1 has 4 leaf nodes
The node 2 has 1 leaf nodes
The node 3 has 3 leaf nodes
The node 4 has 1 leaf nodes
The node 5 has 1 leaf nodes
The node 6 has 1 leaf nodes
方法：想法是在给定的树上执行DFS遍历，并为每个节点保留一个数组leaf []以将叶子节点的数量存储在其下面的子树中。
现在，在沿着树递归的同时，如果找到一个叶子节点，则将其leaf [i]值设置为1并向上返回。现在，每次从函数调用返回向上时，都在其下方添加节点的叶节点。
一旦DFS遍历完成，我们将在数组leaf []中获得叶子节点的数量。
下面是上述方法的实现：
C++
// C++ program to print the number of
// leaf nodes of every node
#include
using namespace std;
// Function to insert edges of tree
void insert(int x, int y, vector adjacency[])
{
adjacency[x].push_back(y);
}
// Function to run DFS on a tree
void dfs(int node, int leaf[], int vis[],
vector adjacency[])
{
leaf[node] = 0;
vis[node] = 1;
// iterate on all the nodes
// connected to node
for (auto it : adjacency[node]) {
// If not visited
if (!vis[it]) {
dfs(it, leaf, vis, adjacency);
leaf[node] += leaf[it];
}
}
if (!adjacency[node].size())
leaf[node] = 1;
}
// Function to print number of
// leaf nodes of a node
void printLeaf(int n, int leaf[])
{
// Function to print leaf nodes
for (int i = 1; i <= n; i++) {
cout << "The node " << i << " has "
<< leaf[i] << " leaf nodes\n";
}
}
// Driver Code
int main()
{
// Given N-ary Tree
/* 1
/ \
2 3
/ | \
4 5 6 */
int N = 6; // no of nodes
vector adjacency[N + 1]; // adjacency list for tree
insert(1, 2, adjacency);
insert(1, 3, adjacency);
insert(3, 4, adjacency);
insert(3, 5, adjacency);
insert(3, 6, adjacency);
int leaf[N + 1]; // Store count of leaf in subtree of i
int vis[N + 1] = { 0 }; // mark nodes visited
dfs(1, leaf, vis, adjacency);
printLeaf(N, leaf);
return 0;
} Java
// Java program to print the number of
// leaf nodes of every node
import java.util.*;
class GFG
{
static Vector> adjacency = new
Vector>();
// Function to insert edges of tree
static void insert(int x, int y)
{
adjacency.get(x).add(y);
}
// Function to run DFS on a tree
static void dfs(int node, int leaf[], int vis[])
{
leaf[node] = 0;
vis[node] = 1;
// iterate on all the nodes
// connected to node
for (int i = 0; i < adjacency.get(node).size(); i++)
{
int it = adjacency.get(node).get(i);
// If not visited
if (vis[it] == 0)
{
dfs(it, leaf, vis);
leaf[node] += leaf[it];
}
}
if (adjacency.get(node).size() == 0)
leaf[node] = 1;
}
// Function to print number of
// leaf nodes of a node
static void printLeaf(int n, int leaf[])
{
// Function to print leaf nodes
for (int i = 1; i <= n; i++)
{
System.out.print( "The node " + i + " has " +
leaf[i] + " leaf nodes\n");
}
}
// Driver Code
public static void main(String args[])
{
// Given N-ary Tree
/* 1
/ \
2 3
/ | \
4 5 6 */
int N = 6; // no of nodes
for(int i = 0; i <= N; i++)
adjacency.add(new Vector());
insert(1, 2);
insert(1, 3);
insert(3, 4);
insert(3, 5);
insert(3, 6);
// Store count of leaf in subtree of i
int leaf[] = new int[N + 1];
// mark nodes visited
int vis[] = new int[N + 1] ;
dfs(1, leaf, vis);
printLeaf(N, leaf);
}
}
// This code is contributed by Arnab Kundu Python3
# Python3 program to print the number of
# leaf nodes of every node
adjacency = [[] for i in range(100)]
# Function to insert edges of tree
def insert(x, y):
adjacency[x].append(y)
# Function to run DFS on a tree
def dfs(node, leaf, vis):
leaf[node] = 0
vis[node] = 1
# iterate on all the nodes
# connected to node
for it in adjacency[node]:
# If not visited
if (vis[it] == False):
dfs(it, leaf, vis)
leaf[node] += leaf[it]
if (len(adjacency[node]) == 0):
leaf[node] = 1
# Function to prnumber of
# leaf nodes of a node
def printLeaf(n, leaf):
# Function to prleaf nodes
for i in range(1, n + 1):
print("The node", i, "has",
leaf[i], "leaf nodes")
# Driver Code
# Given N-ary Tree
'''
/* 1
/ \
2 3
/ | \
4 5 6 '''
N = 6 # no of nodes
# adjacency list for tree
insert(1, 2)
insert(1, 3)
insert(3, 4)
insert(3, 5)
insert(3, 6)
# Store count of leaf in subtree of i
leaf = [0 for i in range(N + 1)]
# mark nodes visited
vis = [0 for i in range(N + 1)]
dfs(1, leaf, vis)
printLeaf(N, leaf)
# This code is contributed by Mohit KumarC#
// C# program to print the number of
// leaf nodes of every node
using System;
using System.Collections.Generic;
class GFG
{
static List> adjacency = new
List>();
// Function to insert edges of tree
static void insert(int x, int y)
{
adjacency[x].Add(y);
}
// Function to run DFS on a tree
static void dfs(int node, int []leaf, int []vis)
{
leaf[node] = 0;
vis[node] = 1;
// iterate on all the nodes
// connected to node
for (int i = 0; i < adjacency[node].Count; i++)
{
int it = adjacency[node][i];
// If not visited
if (vis[it] == 0)
{
dfs(it, leaf, vis);
leaf[node] += leaf[it];
}
}
if (adjacency[node].Count == 0)
leaf[node] = 1;
}
// Function to print number of
// leaf nodes of a node
static void printLeaf(int n, int []leaf)
{
// Function to print leaf nodes
for (int i = 1; i <= n; i++)
{
Console.Write( "The node " + i + " has " +
leaf[i] + " leaf nodes\n");
}
}
// Driver Code
public static void Main(String []args)
{
// Given N-ary Tree
/* 1
/ \
2 3
/ | \
4 5 6 */
int N = 6; // no of nodes
for(int i = 0; i <= N; i++)
adjacency.Add(new List());
insert(1, 2);
insert(1, 3);
insert(3, 4);
insert(3, 5);
insert(3, 6);
// Store count of leaf in subtree of i
int []leaf = new int[N + 1];
// mark nodes visited
int []vis = new int[N + 1] ;
dfs(1, leaf, vis);
printLeaf(N, leaf);
}
}
// This code contributed by Rajput-Ji 输出：
The node 1 has 4 leaf nodes
The node 2 has 1 leaf nodes
The node 3 has 3 leaf nodes
The node 4 has 1 leaf nodes
The node 5 has 1 leaf nodes
The node 6 has 1 leaf nodes
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