给定一棵二叉树，其中节点包含字符，任务是计算从根顶点到叶的路径数，以使路径中节点值的至少一个排列是回文。
例子：
Input:
2
/ \
3 1
/ \ \
3 4 2
/ \ / \
2 1 2 1
Output: 2
Explanation:
Paths whose one of the
permuation are palindrome are -
2 => 3 => 3 => 2 and
2 => 1 => 2 => 1
Input:
2
/ \
a 3
/ \
2 a
Output: 2
Explanation:
Palindromic paths are
2 => a => 2 and
2 => a => a
方法：想法是使用预遍历遍历二叉树并跟踪路径。每当到达叶节点时，请检查当前路径中节点值的任何排列是否是回文路径。
要检查节点值的排列是否回文或不使用映射保持每个字符的频率。如果奇数频率的元素数最多为1，则该路径将是回文路径。
下面是上述方法的实现：
C++
// C++ implementation to count of
// the path whose permutation is
// a palindromic path
#include
using namespace std;
#define ll long long
// Map to store the frequency
map freq;
int ans = 0;
// Structure of the node
struct Node {
char val;
struct Node *left, *right;
};
// Function to add new node
Node* newNode(char key)
{
Node* temp = new Node;
temp->val = key;
temp->left = temp->right = NULL;
return (temp);
}
// Function to check that the path
// is a palindrome or not
int checkPalin()
{
int oddCount = 0;
for (auto x : freq) {
if (x.second % 2 == 1)
oddCount++;
}
return oddCount <= 1;
}
// Function to count the root to
// leaf path whose permutation is
// a palindromic path
void cntpalin(Node* root)
{
if (root == NULL)
return;
freq[root->val]++;
if (root->left == NULL
&& root->right == NULL) {
if (checkPalin() == true)
ans++;
}
cntpalin(root->left);
cntpalin(root->right);
freq[root->val]--;
}
// Driver Code
int main()
{
Node* root = newNode('2');
root->left = newNode('a');
root->left->right = newNode('a');
root->left->left = newNode('2');
root->left->right->right = newNode('2');
root->right = newNode('3');
// Function Call
cntpalin(root);
cout << ans << endl;
return 0;
} Java
// Java implementation to count of
// the path whose permutation is
// a palindromic path
import java.util.*;
class GFG{
// Map to store the frequency
static HashMap freq =
new HashMap<>();
static int ans = 0;
// Structure of the node
static class Node
{
char val;
Node left, right;
};
// Function to add new node
static Node newNode(char key)
{
Node temp = new Node();
temp.val = key;
temp.left = temp.right = null;
return (temp);
}
// Function to check that the path
// is a palindrome or not
static boolean checkPalin()
{
int oddCount = 0;
for (Map.Entry x : freq.entrySet())
{
if (x.getValue() % 2 == 1)
oddCount++;
}
return oddCount <= 1 ? true : false;
}
// Function to count the root to
// leaf path whose permutation is
// a palindromic path
static void cntpalin(Node root)
{
if (root == null)
return;
if(freq.containsKey(root.val))
{
freq.put(root.val,
freq.get(root.val) + 1);
}
else
{
freq.put(root.val, 1);
}
if (root.left == null &&
root.right == null)
{
if (checkPalin() == true)
ans++;
}
cntpalin(root.left);
cntpalin(root.right);
if(freq.containsKey(root.val))
{
freq.put(root.val,
freq.get(root.val) - 1);
}
}
// Driver Code
public static void main(String[] args)
{
Node root = newNode('2');
root.left = newNode('a');
root.left.right = newNode('a');
root.left.left = newNode('2');
root.left.right.right = newNode('2');
root.right = newNode('3');
// Function Call
cntpalin(root);
System.out.print(ans + "\n");
}
}
// This code is contributed by Princi Singh Python3
# Python3 program for the
# above approach
from collections import deque
# A Tree node
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
freq = {}
ans = 0
# Function to check that the path
# is a palindrome or not
def checkPalin():
oddCount = 0
for x in freq:
if (freq[x] % 2 == 1):
oddCount+=1
return oddCount <= 1
# Function to count the root to
# leaf path whose permutation is
# a palindromic path
def cntpalin(root):
global freq, ans
if (root == None):
return
freq[root.data] = freq.get(root.data,
0) + 1
if (root.left == None
and root.right == None):
if (checkPalin() == True):
ans += 1
cntpalin(root.left)
cntpalin(root.right)
freq[root.data] -= 1
# Driver Code
if __name__ == '__main__':
root = Node('2')
root.left = Node('a')
root.left.right = Node('a')
root.left.left = Node('2')
root.left.right.right = Node('2')
root.right = Node('3')
# Function Call
cntpalin(root)
print(ans)
# This code is contributed by Rutvik_56C#
// C# implementation to count of
// the path whose permutation is
// a palindromic path
using System;
using System.Collections.Generic;
class GFG{
// Map to store the frequency
static Dictionary freq = new Dictionary();
static int ans = 0;
// Structure of the node
public class Node
{
public char val;
public Node left,
right;
};
// Function to add new node
static Node newNode(char key)
{
Node temp = new Node();
temp.val = key;
temp.left = temp.right = null;
return (temp);
}
// Function to check that
// the path is a palindrome
// or not
static bool checkPalin()
{
int oddCount = 0;
foreach (KeyValuePair x in freq)
{
if (x.Value % 2 == 1)
oddCount++;
}
return oddCount <= 1 ? true : false;
}
// Function to count the root to
// leaf path whose permutation is
// a palindromic path
static void cntpalin(Node root)
{
if (root == null)
return;
if(freq.ContainsKey(root.val))
{
freq[root.val] = freq[root.val] + 1;
}
else
{
freq.Add(root.val, 1);
}
if (root.left == null &&
root.right == null)
{
if (checkPalin() == true)
ans++;
}
cntpalin(root.left);
cntpalin(root.right);
if(freq.ContainsKey(root.val))
{
freq[root.val] = freq[root.val] - 1;
}
}
// Driver Code
public static void Main(String[] args)
{
Node root = newNode('2');
root.left = newNode('a');
root.left.right = newNode('a');
root.left.left = newNode('2');
root.left.right.right = newNode('2');
root.right = newNode('3');
// Function Call
cntpalin(root);
Console.Write(ans + "\n");
}
}
// This code is contributed by Amit Katiyar 输出：
2
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