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📜  在二叉树中排列为回文的根到叶路径的计数

📅  最后修改于: 2021-09-04 11:23:07             🧑  作者: Mango

给定一棵节点包含字符的二叉树,任务是计算从根顶点到叶子的路径数,使得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
permutation 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 

做法:思路是使用先序遍历来遍历二叉树,并跟踪路径。每当到达叶节点时,检查当前路径中节点值的任何排列是否是回文路径。
检查节点值的排列是回文还是不使用映射维护每个字符的频率。如果奇数频率元素的数量最多为1,则路径将是回文的。
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_56


C#
// 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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