给定一个N 叉树，以及所有节点的字符串形式的权重，任务是计算权重为回文的叶节点的数量。
例子：
Input:
1(ab)
/ \
(abca)2 5 (aba)
/ \
(axxa)3 4 (geeks)
Output: 2
Explanation:
Only the weights of the leaf nodes
"axxa" and "aba" are palindromes.
Input:
1(abx)
/
2(abaa)
/
3(amma)
Output: 1
Explanation:
Only the weight of the leaf
node "amma" is palindrome.
方法：解决上述问题，请按照以下步骤操作：
- 深度优先搜索可用于遍历整个树。
- 我们将在遍历时跟踪父节点以避免访问的节点数组。
- 最初，我们可以为每个节点设置一个标志，如果该节点至少有一个子节点(即非叶节点)，那么我们将重置该标志。
- 没有孩子的节点是叶节点。对于每个叶节点，我们将检查它的字符串是否为回文。如果是，则增加计数。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
int cnt = 0;
vector graph[100];
vector weight(100);
// Function that returns true
// if x is a palindrome
bool isPalindrome(string x)
{
int n = x.size();
for (int i = 0; i < n / 2; i++) {
if (x[i] != x[n - 1 - i])
return false;
}
return true;
}
// Function to perform DFS on the tree
void dfs(int node, int parent)
{
int flag = 1;
// Iterating the children of current node
for (int to : graph[node]) {
// There is at least a child
// of the current node
if (to == parent)
continue;
flag = 0;
dfs(to, node);
}
// Current node is connected to only
// its parent i.e. it is a leaf node
if (flag == 1) {
// Weight of the current node
string x = weight[node];
// If the weight is a palindrome
if (isPalindrome(x))
cnt += 1;
}
}
// Driver code
int main()
{
// Weights of the node
weight[1] = "ab";
weight[2] = "abca";
weight[3] = "axxa";
weight[4] = "geeks";
weight[5] = "aba";
// Edges of the tree
graph[1].push_back(2);
graph[2].push_back(3);
graph[2].push_back(4);
graph[1].push_back(5);
dfs(1, 1);
cout << cnt;
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG{
static int cnt = 0;
static Vector []graph = new Vector[100];
static String []weight = new String[100];
// Function that returns true
// if x is a palindrome
static boolean isPalindrome(String x)
{
int n = x.length();
for (int i = 0; i < n / 2; i++)
{
if (x.charAt(i) != x.charAt(n - 1 - i))
return false;
}
return true;
}
// Function to perform DFS on the tree
static void dfs(int node, int parent)
{
int flag = 1;
// Iterating the children of current node
for (int to : graph[node])
{
// There is at least a child
// of the current node
if (to == parent)
continue;
flag = 0;
dfs(to, node);
}
// Current node is connected to only
// its parent i.e. it is a leaf node
if (flag == 1)
{
// Weight of the current node
String x = weight[node];
// If the weight is a palindrome
if (isPalindrome(x))
cnt += 1;
}
}
// Driver code
public static void main(String[] args)
{
for(int i = 0; i < graph.length;i++)
graph[i] = new Vector();
// Weights of the node
weight[1] = "ab";
weight[2] = "abca";
weight[3] = "axxa";
weight[4] = "geeks";
weight[5] = "aba";
// Edges of the tree
graph[1].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
dfs(1, 1);
System.out.print(cnt);
}
}
// This code is contributed by amal kumar choubey Python3
# Python3 implementation of the approach
cnt = 0
graph = [0] * 100
for i in range(100):
graph[i] = []
weight = [0] * 100
# Function that returns true
# if x is a palindrome
def isPalindrome(x: str) -> bool:
n = len(x)
for i in range(n // 2):
if (x[i] != x[n - 1 - i]):
return False
return True
# Function to perform DFS on the tree
def dfs(node: int, parent: int) -> None:
global cnt, graph, weight
flag = 1
# Iterating the children of current node
for to in graph[node]:
# There is at least a child
# of the current node
if (to == parent):
continue
flag = 0
dfs(to, node)
# Current node is connected to only
# its parent i.e. it is a leaf node
if (flag == 1):
# Weight of the current node
x = weight[node]
# If the weight is a palindrome
if (isPalindrome(x)):
cnt += 1
# Driver code
if __name__ == "__main__":
# Weights of the node
weight[1] = "ab"
weight[2] = "abca"
weight[3] = "axxa"
weight[4] = "geeks"
weight[5] = "aba"
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
dfs(1, 1)
print(cnt)
# This code is contributed by sanjeev2552C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
static int cnt = 0;
static List []graph = new List[100];
static String []weight = new String[100];
// Function that returns true
// if x is a palindrome
static bool isPalindrome(String x)
{
int n = x.Length;
for(int i = 0; i < n / 2; i++)
{
if (x[i] != x[n - 1 - i])
return false;
}
return true;
}
// Function to perform DFS on the tree
static void dfs(int node, int parent)
{
int flag = 1;
// Iterating the children of
// current node
foreach (int to in graph[node])
{
// There is at least a child
// of the current node
if (to == parent)
continue;
flag = 0;
dfs(to, node);
}
// Current node is connected to only
// its parent i.e. it is a leaf node
if (flag == 1)
{
// Weight of the current node
String x = weight[node];
// If the weight is a palindrome
if (isPalindrome(x))
cnt += 1;
}
}
// Driver code
public static void Main(String[] args)
{
for(int i = 0; i < graph.Length; i++)
graph[i] = new List();
// Weights of the node
weight[1] = "ab";
weight[2] = "abca";
weight[3] = "axxa";
weight[4] = "geeks";
weight[5] = "aba";
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
Console.Write(cnt);
}
}
// This code is contributed by amal kumar choubey 输出：
2
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