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Count of root to leaf paths whose permutation is palindrome in a Binary Tree

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  • Difficulty Level : Easy
  • Last Updated : 19 Sep, 2022
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Given a binary tree where node contains characters, the task is to count the number of paths from root vertex to leaf such that at least one permutation of the node values in the path is a palindrome.
Examples: 

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 

Approach: The idea is to use pre-order traversal to traverse the binary tree and keep track of the path. Whenever a leaf node is reached then check that if any permutation of nodes values in the current path is a palindromic path or not.
To check the permutation of the values of the nodes is palindromic or not maintain the frequency of each character using a map. The path will be palindromic if the number of elements with odd frequency is at most 1. 
Below is the implementation of the above approach: 
 

C++




// C++ implementation to count of
// the path whose permutation is
// a palindromic path
 
#include <bits/stdc++.h>
using namespace std;
#define ll long long
 
// Map to store the frequency
map<char, int> 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<Character, Integer> 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<Character, Integer> 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<char, int> freq
        = new Dictionary<char, int>();
    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<char, int> 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


Javascript




<script>
// Javascript implementation to count of
// the path whose permutation is
// a palindromic path
 
// Map to store the frequency
let  freq = new Map();
 
let ans = 0;
 
// Structure of the node
class Node
{
    constructor(key)
    {
        this.val = key;
        this.left = this.right = null;
    }
}
 
// Function to check that the path
// is a palindrome or not
function checkPalin()
{
    let oddCount = 0;
  for (let [key, value] of freq.entries())
  {
    if (value % 2 == 1)
      oddCount++;
  }
  
  return oddCount <= 1 ? true : false;
}
 
// Function to count the root to
// leaf path whose permutation is
// a palindromic path
function cntpalin(root)
{
    if (root == null)
    return;
  if(freq.has(root.val))
  {
    freq.set(root.val,
    freq.get(root.val) + 1);
  }
  else
  {
    freq.set(root.val, 1);
  }
  
  if (root.left == null &&
      root.right == null)
  {
    if (checkPalin() == true)
      ans++;
  }
    
  cntpalin(root.left);
  cntpalin(root.right);
    
  if(freq.has(root.val))
  {
    freq.set(root.val,
    freq.get(root.val) - 1);
  }
}
 
// Driver Code
let root = new Node('2');
root.left = new Node('a');
root.left.right = new Node('a');
root.left.left = new Node('2');
root.left.right.right = new Node('2');
root.right = new Node('3');
 
// Function Call
cntpalin(root);
 
document.write(ans + "<br>");
 
// This code is contributed by unknown2108.
</script>


Output: 

2

 

Time Complexity: O(N*(2h)), where N is the number of nodes in the binary tree and h is the height. 

Space Complexity: O(N), where N is the number of nodes in the binary tree.


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