Count tree paths whose permutation is palindrome

Here given code implementation process.

import java.util.HashMap;
/*
  Java program
  Count tree paths whose permutation is palindrome
*/
// Binary Tree node
class TreeNode
{
	public char data;
	public TreeNode left;
	public TreeNode right;
	public TreeNode(char data)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
public class BinaryTree
{
	public TreeNode root;
	public int result;
	public BinaryTree()
	{
		// Set initial tree root to null
		this.root = null;
		this.result = 0;
	}
	public boolean isPalindrome(HashMap < Character, Integer > record)
	{
		boolean odd = false;
		// Check that if given string can be palindrome
		for (char info: record.keySet())
		{
			if (record.get(info) % 2 != 0)
			{
				if (odd == false)
				{
					// One odd allow
					odd = true;
				}
				else
				{
					// When have more than two odd character
					return false;
				}
			}
		}
		return true;
	}
	public void getPath(TreeNode node, HashMap < Character, Integer > record)
	{
		if (node == null)
		{
			return;
		}
		if (record.containsKey(node.data))
		{
			// Increase node frequency
			record.put(node.data, record.get(node.data) + 1);
		}
		else
		{
			// Add new node in record
			record.put(node.data, 1);
		}
		// Visit left subtree
		getPath(node.left, record);
		// Visit right subtree
		getPath(node.right, record);
		if (node.left == null && node.right == null)
		{
			// When node is leaf node
			if (isPalindrome(record))
			{
				// Increase the resultant counter
				// When path are generated a palindrome
				this.result++;
			}
		}
		if (record.get(node.data) == 1)
		{
			// Remove the node from  record
			record.remove(node.data);
		}
		else
		{
			record.put(node.data, record.get(node.data) - 1);
		}
	}
	// Display inorder sequence in binary tree
	public void printInorder(TreeNode node)
	{
		if (node == null)
		{
			return;
		}
		// Visit left subtree
		printInorder(node.left);
		// print node value
		System.out.print("  " + node.data);
		// Visit right subtree
		printInorder(node.right);
	}
	public void palindromePath()
	{
		//Use to count frequency of path elements
		HashMap < Character, Integer > record = 
          new HashMap < Character, Integer > ();
		this.result = 0;
		getPath(this.root, record);
		// Display inorder of binary tree
		printInorder(this.root);
		// Display calculated result
		System.out.println("\n Permutable palindromic paths is " + this.result);
	}
	public static void main(String[] args)
	{
		// Create new binary tree
		BinaryTree tree1 = new BinaryTree();
		BinaryTree tree2 = new BinaryTree();
		/*
		     a   
		    / \                         
		   /   \    
		  c     e    
		 / \     \               
		a   a     a
		   / \   / \
		  d   c e   b
		-----------------
		Constructing binary tree   
		*/
		tree1.root = new TreeNode('a');
		tree1.root.left = new TreeNode('c');
		tree1.root.left.right = new TreeNode('a');
		tree1.root.left.right.left = new TreeNode('d');
		tree1.root.left.right.right = new TreeNode('c');
		tree1.root.left.left = new TreeNode('a');
		tree1.root.right = new TreeNode('e');
		tree1.root.right.right = new TreeNode('a');
		tree1.root.right.right.right = new TreeNode('b');
		tree1.root.right.right.left = new TreeNode('e');
		// Test A
        // a  c  a
        // a  c  c a  or c  a  a  c  
        // a  e  e a  or e  a  a  e
        // Result : 3
		tree1.palindromePath();
		/*
		     a  
		    / \                          
		   /   \    
		  a     h    
		   \     \               
		    a     e
		   / \   / \
		  d   a b   h
		             \
		              e
		-----------------
		Constructing binary tree
		*/
		tree2.root = new TreeNode('a');
		tree2.root.left = new TreeNode('a');
		tree2.root.left.right = new TreeNode('a');
		tree2.root.left.right.left = new TreeNode('d');
		tree2.root.left.right.right = new TreeNode('a');
		tree2.root.right = new TreeNode('h');
		tree2.root.right.right = new TreeNode('e');
		tree2.root.right.right.right = new TreeNode('h');
		tree2.root.right.right.left = new TreeNode('b');
		tree2.root.right.right.right.right = new TreeNode('e');
		// Test B
      	// [a  a  a  a]
        // [h  e  a  e  h] or [e h a h e]
        // Result : 2
		tree2.palindromePath();
	}
}

input

  a  c  d  a  c  a  e  e  a  b
 Permutable palindromic paths is 3
  a  d  a  a  a  h  b  e  h  e
 Permutable palindromic paths is 2
// Include header file
#include <iostream>
#include <unordered_map>
using namespace std;

/*
  C++ program
  Count tree paths whose permutation is palindrome
*/
// Binary Tree node
class TreeNode
{
	public: 
    char data;
	TreeNode *left;
	TreeNode *right;
	TreeNode(char data)
	{
		// Set node value
		this->data = data;
		this->left = NULL;
		this->right = NULL;
	}
};
class BinaryTree
{
	public: TreeNode *root;
	int result;
	BinaryTree()
	{
		this->root = NULL;
		this->result = 0;
	}
	bool isPalindrome(unordered_map < char, int > record)
	{
		bool odd = false;
		// Check that if given string can be palindrome
		for (auto &info: record)
		{
			if (info.second % 2 != 0)
			{
				if (odd == false)
				{
					// One odd allow
					odd = true;
				}
				else
				{
					// When have more than two odd character
					return false;
				}
			}
		}
		return true;
	}
	void getPath(TreeNode *node, unordered_map < char, int > &record)
	{
		if (node == NULL)
		{
			return;
		}
		if (record.find(node->data) != record.end())
		{
			// Increase node frequency
			record[node->data] = record[node->data] + 1;
		}
		else
		{
			// Add new node in record
			record[node->data] = 1;
		}
		// Visit left subtree
		this->getPath(node->left, record);
		// Visit right subtree
		this->getPath(node->right, record);
		if (node->left == NULL && node->right == NULL)
		{
			// When node is leaf node
			if (this->isPalindrome(record))
			{
				// Increase the resultant counter
				// When path are generated a palindrome
				this->result++;
			}
		}
		if (record[node->data] == 1)
		{
			// Remove the node from  record
			record.erase(node->data);
		}
		else
		{
			record[node->data] = record[node->data] - 1;
		}
	}
	// Display inorder sequence in binary tree
	void printInorder(TreeNode *node)
	{
		if (node == NULL)
		{
			return;
		}
		// Visit left subtree
		this->printInorder(node->left);
		// print node value
		cout << "  " << node->data;
		// Visit right subtree
		this->printInorder(node->right);
	}
	void palindromePath()
	{
		//Use to count frequency of path elements
		unordered_map < char, int > record ;
		this->result = 0;
		this->getPath(this->root, record);
		// Display inorder of binary tree
		this->printInorder(this->root);
		// Display calculated result
		cout << "\n Permutable palindromic paths is " << this->result << endl;
	}
};
int main()
{
	// Create new binary tree
	BinaryTree *tree1 = new BinaryTree();
	BinaryTree *tree2 = new BinaryTree();
	/*
	     a   
	    / \                         
	   /   \    
	  c     e    
	 / \     \               
	a   a     a
	   / \   / \
	  d   c e   b
	-----------------
	Constructing binary tree   
	*/
	tree1->root = new TreeNode('a');
	tree1->root->left = new TreeNode('c');
	tree1->root->left->right = new TreeNode('a');
	tree1->root->left->right->left = new TreeNode('d');
	tree1->root->left->right->right = new TreeNode('c');
	tree1->root->left->left = new TreeNode('a');
	tree1->root->right = new TreeNode('e');
	tree1->root->right->right = new TreeNode('a');
	tree1->root->right->right->right = new TreeNode('b');
	tree1->root->right->right->left = new TreeNode('e');
	// Test A
	// a  c  a
	// a  c  c a  or c  a  a  c
	// a  e  e a  or e  a  a  e
	// Result : 3
	tree1->palindromePath();
	/*
	     a  
	    / \                          
	   /   \    
	  a     h    
	   \     \               
	    a     e
	   / \   / \
	  d   a b   h
	             \
	              e
	-----------------
	Constructing binary tree
	*/
	tree2->root = new TreeNode('a');
	tree2->root->left = new TreeNode('a');
	tree2->root->left->right = new TreeNode('a');
	tree2->root->left->right->left = new TreeNode('d');
	tree2->root->left->right->right = new TreeNode('a');
	tree2->root->right = new TreeNode('h');
	tree2->root->right->right = new TreeNode('e');
	tree2->root->right->right->right = new TreeNode('h');
	tree2->root->right->right->left = new TreeNode('b');
	tree2->root->right->right->right->right = new TreeNode('e');
	// Test B
	// [a  a  a  a]
	// [h  e  a  e  h] or [e h a h e]
	// Result : 2
	tree2->palindromePath();
	return 0;
}

input

  a  c  d  a  c  a  e  e  a  b
 Permutable palindromic paths is 3
  a  d  a  a  a  h  b  e  h  e
 Permutable palindromic paths is 2
// Include namespace system
using System;
using System.Collections.Generic;
/*
  Csharp program
  Count tree paths whose permutation is palindrome
*/
// Binary Tree node
public class TreeNode
{
	public char data;
	public TreeNode left;
	public TreeNode right;
	public TreeNode(char data)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
public class BinaryTree
{
	public TreeNode root;
	public int result;
	public BinaryTree()
	{
		// Set initial tree root to null
		this.root = null;
		this.result = 0;
	}
	public Boolean isPalindrome(Dictionary < char, int > record)
	{
		Boolean odd = false;
		// Check that if given string can be palindrome
		foreach(KeyValuePair < char, int > info in record)
		{
			if (info.Value % 2 != 0)
			{
				if (odd == false)
				{
					// One odd allow
					odd = true;
				}
				else
				{
					// When have more than two odd character
					return false;
				}
			}
		}
		return true;
	}
	public void getPath(TreeNode node, Dictionary < char, int > record)
	{
		if (node == null)
		{
			return;
		}
		if (record.ContainsKey(node.data))
		{
			// Increase node frequency
			record[node.data] = record[node.data] + 1;
		}
		else
		{
			// Add new node in record
			record.Add(node.data, 1);
		}
		// Visit left subtree
		this.getPath(node.left, record);
		// Visit right subtree
		this.getPath(node.right, record);
		if (node.left == null && node.right == null)
		{
			// When node is leaf node
			if (this.isPalindrome(record))
			{
				// Increase the resultant counter
				// When path are generated a palindrome
				this.result++;
			}
		}
		if (record[node.data] == 1)
		{
			// Remove the node from  record
			record.Remove(node.data);
		}
		else
		{
			record[node.data] = record[node.data] - 1;
		}
	}
	// Display inorder sequence in binary tree
	public void printInorder(TreeNode node)
	{
		if (node == null)
		{
			return;
		}
		// Visit left subtree
		this.printInorder(node.left);
		// print node value
		Console.Write("  " + node.data);
		// Visit right subtree
		this.printInorder(node.right);
	}
	public void palindromePath()
	{
		//Use to count frequency of path elements
		Dictionary < char, int > record = new Dictionary < char, int > ();
		this.result = 0;
		this.getPath(this.root, record);
		// Display inorder of binary tree
		this.printInorder(this.root);
		// Display calculated result
		Console.WriteLine("\n Permutable palindromic paths is " + this.result);
	}
	public static void Main(String[] args)
	{
		// Create new binary tree
		BinaryTree tree1 = new BinaryTree();
		BinaryTree tree2 = new BinaryTree();
		/*
		     a   
		    / \                         
		   /   \    
		  c     e    
		 / \     \               
		a   a     a
		   / \   / \
		  d   c e   b
		-----------------
		Constructing binary tree   
		*/
		tree1.root = new TreeNode('a');
		tree1.root.left = new TreeNode('c');
		tree1.root.left.right = new TreeNode('a');
		tree1.root.left.right.left = new TreeNode('d');
		tree1.root.left.right.right = new TreeNode('c');
		tree1.root.left.left = new TreeNode('a');
		tree1.root.right = new TreeNode('e');
		tree1.root.right.right = new TreeNode('a');
		tree1.root.right.right.right = new TreeNode('b');
		tree1.root.right.right.left = new TreeNode('e');
		// Test A
		// a  c  a
		// a  c  c a  or c  a  a  c
		// a  e  e a  or e  a  a  e
		// Result : 3
		tree1.palindromePath();
		/*
		     a  
		    / \                          
		   /   \    
		  a     h    
		   \     \               
		    a     e
		   / \   / \
		  d   a b   h
		             \
		              e
		-----------------
		Constructing binary tree
		*/
		tree2.root = new TreeNode('a');
		tree2.root.left = new TreeNode('a');
		tree2.root.left.right = new TreeNode('a');
		tree2.root.left.right.left = new TreeNode('d');
		tree2.root.left.right.right = new TreeNode('a');
		tree2.root.right = new TreeNode('h');
		tree2.root.right.right = new TreeNode('e');
		tree2.root.right.right.right = new TreeNode('h');
		tree2.root.right.right.left = new TreeNode('b');
		tree2.root.right.right.right.right = new TreeNode('e');
		// Test B
		// [a  a  a  a]
		// [h  e  a  e  h] or [e h a h e]
		// Result : 2
		tree2.palindromePath();
	}
}

input

  a  c  d  a  c  a  e  e  a  b
 Permutable palindromic paths is 3
  a  d  a  a  a  h  b  e  h  e
 Permutable palindromic paths is 2
<?php
/*
  Php program
  Count tree paths whose permutation is palindrome
*/
// Binary Tree node
class TreeNode
{
	public $data;
	public $left;
	public $right;
	public	function __construct($data)
	{
		// Set node value
		$this->data = $data;
		$this->left = NULL;
		$this->right = NULL;
	}
}
class BinaryTree
{
	public $root;
	public $result;
	public	function __construct()
	{
		$this->root = NULL;
		$this->result = 0;
	}
	public	function isPalindrome($record)
	{
		$odd = false;
		// Check that if given string can be palindrome
		foreach($record as $key => $value)
		{
			if ($value % 2 != 0)
			{
				if ($odd == false)
				{
					// One odd allow
					$odd = true;
				}
				else
				{
					// When have more than two odd character
					return false;
				}
			}
		}
		return true;
	}
	public	function getPath($node, &$record)
	{
		if ($node == NULL)
		{
			return;
		}
		if (array_key_exists($node->data, $record))
		{
			// Increase node frequency
			$record[$node->data] = $record[$node->data] + 1;
		}
		else
		{
			// Add new node in record
			$record[$node->data] = 1;
		}
		// Visit left subtree
		$this->getPath($node->left, $record);
		// Visit right subtree
		$this->getPath($node->right, $record);
		if ($node->left == NULL && $node->right == NULL)
		{
			// When node is leaf node
			if ($this->isPalindrome($record))
			{
				// Increase the resultant counter
				// When path are generated a palindrome
				$this->result++;
			}
		}
		if ($record[$node->data] == 1)
		{
			// Remove the node from  record
			unset($record[$node->data]);
		}
		else
		{
			$record[$node->data] = $record[$node->data] - 1;
		}
	}
	// Display inorder sequence in binary tree
	public	function printInorder($node)
	{
		if ($node == NULL)
		{
			return;
		}
		// Visit left subtree
		$this->printInorder($node->left);
		// print node value
		echo("  ".$node->data);
		// Visit right subtree
		$this->printInorder($node->right);
	}
	public	function palindromePath()
	{
		//Use to count frequency of path elements
		$record = array();
		$this->result = 0;
		$this->getPath($this->root, $record);
		// Display inorder of binary tree
		$this->printInorder($this->root);
		// Display calculated result
		echo("\n Permutable palindromic paths is ".$this->result."\n");
	}
}

function main()
{
	// Create new binary tree
	$tree1 = new BinaryTree();
	$tree2 = new BinaryTree();
	/*
	     a   
	    / \                         
	   /   \    
	  c     e    
	 / \     \               
	a   a     a
	   / \   / \
	  d   c e   b
	-----------------
	Constructing binary tree   
	*/
	$tree1->root = new TreeNode('a');
	$tree1->root->left = new TreeNode('c');
	$tree1->root->left->right = new TreeNode('a');
	$tree1->root->left->right->left = new TreeNode('d');
	$tree1->root->left->right->right = new TreeNode('c');
	$tree1->root->left->left = new TreeNode('a');
	$tree1->root->right = new TreeNode('e');
	$tree1->root->right->right = new TreeNode('a');
	$tree1->root->right->right->right = new TreeNode('b');
	$tree1->root->right->right->left = new TreeNode('e');
	// Test A
	// a  c  a
	// a  c  c a  or c  a  a  c
	// a  e  e a  or e  a  a  e
	// Result : 3
	$tree1->palindromePath();
	/*
	     a  
	    / \                          
	   /   \    
	  a     h    
	   \     \               
	    a     e
	   / \   / \
	  d   a b   h
	             \
	              e
	-----------------
	Constructing binary tree
	*/
	$tree2->root = new TreeNode('a');
	$tree2->root->left = new TreeNode('a');
	$tree2->root->left->right = new TreeNode('a');
	$tree2->root->left->right->left = new TreeNode('d');
	$tree2->root->left->right->right = new TreeNode('a');
	$tree2->root->right = new TreeNode('h');
	$tree2->root->right->right = new TreeNode('e');
	$tree2->root->right->right->right = new TreeNode('h');
	$tree2->root->right->right->left = new TreeNode('b');
	$tree2->root->right->right->right->right = new TreeNode('e');
	// Test B
	// [a  a  a  a]
	// [h  e  a  e  h] or [e h a h e]
	// Result : 2
	$tree2->palindromePath();
}
main();

input

  a  c  d  a  c  a  e  e  a  b
 Permutable palindromic paths is 3
  a  d  a  a  a  h  b  e  h  e
 Permutable palindromic paths is 2
/*
  Node JS program
  Count tree paths whose permutation is palindrome
*/
// Binary Tree node
class TreeNode
{
	constructor(data)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	constructor()
	{
		this.root = null;
		this.result = 0;
	}
	isPalindrome(record)
	{
		var odd = false;
		// Check that if given string can be palindrome
		for (let [key, value] of record)
		{
			if (value % 2 != 0)
			{
				if (odd == false)
				{
					// One odd allow
					odd = true;
				}
				else
				{
					// When have more than two odd character
					return false;
				}
			}
		}
		return true;
	}
	getPath(node, record)
	{
		if (node == null)
		{
			return;
		}
		if (record.has(node.data))
		{
			// Increase node frequency
			record.set(node.data, record.get(node.data) + 1);
		}
		else
		{
			// Add new node in record
			record.set(node.data, 1);
		}
		// Visit left subtree
		this.getPath(node.left, record);
		// Visit right subtree
		this.getPath(node.right, record);
		if (node.left == null && node.right == null)
		{
			// When node is leaf node
			if (this.isPalindrome(record))
			{
				// Increase the resultant counter
				// When path are generated a palindrome
				this.result++;
			}
		}
		if (record.get(node.data) == 1)
		{
			// Remove the node from  record
			record.delete(node.data);
		}
		else
		{
			record.set(node.data, record.get(node.data) - 1);
		}
	}
	// Display inorder sequence in binary tree
	printInorder(node)
	{
		if (node == null)
		{
			return;
		}
		// Visit left subtree
		this.printInorder(node.left);
		// print node value
		process.stdout.write("  " + node.data);
		// Visit right subtree
		this.printInorder(node.right);
	}
	palindromePath()
	{
		//Use to count frequency of path elements
		var record = new Map();
		this.result = 0;
		this.getPath(this.root, record);
		// Display inorder of binary tree
		this.printInorder(this.root);
		// Display calculated result
		console.log("\n Permutable palindromic paths is " + this.result);
	}
}

function main()
{
	// Create new binary tree
	var tree1 = new BinaryTree();
	var tree2 = new BinaryTree();
	/*
	     a   
	    / \                         
	   /   \    
	  c     e    
	 / \     \               
	a   a     a
	   / \   / \
	  d   c e   b
	-----------------
	Constructing binary tree   
	*/
	tree1.root = new TreeNode('a');
	tree1.root.left = new TreeNode('c');
	tree1.root.left.right = new TreeNode('a');
	tree1.root.left.right.left = new TreeNode('d');
	tree1.root.left.right.right = new TreeNode('c');
	tree1.root.left.left = new TreeNode('a');
	tree1.root.right = new TreeNode('e');
	tree1.root.right.right = new TreeNode('a');
	tree1.root.right.right.right = new TreeNode('b');
	tree1.root.right.right.left = new TreeNode('e');
	// Test A
	// a  c  a
	// a  c  c a  or c  a  a  c
	// a  e  e a  or e  a  a  e
	// Result : 3
	tree1.palindromePath();
	/*
	     a  
	    / \                          
	   /   \    
	  a     h    
	   \     \               
	    a     e
	   / \   / \
	  d   a b   h
	             \
	              e
	-----------------
	Constructing binary tree
	*/
	tree2.root = new TreeNode('a');
	tree2.root.left = new TreeNode('a');
	tree2.root.left.right = new TreeNode('a');
	tree2.root.left.right.left = new TreeNode('d');
	tree2.root.left.right.right = new TreeNode('a');
	tree2.root.right = new TreeNode('h');
	tree2.root.right.right = new TreeNode('e');
	tree2.root.right.right.right = new TreeNode('h');
	tree2.root.right.right.left = new TreeNode('b');
	tree2.root.right.right.right.right = new TreeNode('e');
	// Test B
	// [a  a  a  a]
	// [h  e  a  e  h] or [e h a h e]
	// Result : 2
	tree2.palindromePath();
}
main();

input

  a  c  d  a  c  a  e  e  a  b
 Permutable palindromic paths is 3
  a  d  a  a  a  h  b  e  h  e
 Permutable palindromic paths is 2
#  Python 3 program
#  Count tree paths whose permutation is palindrome

#  Binary Tree node
class TreeNode :
	def __init__(self, data) :
		#  Set node value
		self.data = data
		self.left = None
		self.right = None
	

class BinaryTree :
	def __init__(self) :
		self.root = None
		self.result = 0
	
	def isPalindrome(self, record) :
		odd = False
		#  Check that if given string can be palindrome
		for key, value in record.items() :
			if (value % 2 != 0) :
				if (odd == False) :
					#  One odd allow
					odd = True
				else :
					#  When have more than two odd character
					return False
				
			
		
		return True
	
	def getPath(self, node, record) :
		if (node == None) :
			return
		
		if (node.data in record.keys()) :
			#  Increase node frequency
			record[node.data] = record.get(node.data) + 1
		else :
			#  Add new node in record
			record[node.data] = 1
		
		#  Visit left subtree
		self.getPath(node.left, record)
		#  Visit right subtree
		self.getPath(node.right, record)
		if (node.left == None and node.right == None) :
			#  When node is leaf node
			if (self.isPalindrome(record)) :
				#  Increase the resultant counter
				#  When path are generated a palindrome
				self.result += 1
			
		
		if (record.get(node.data) == 1) :
			#  Remove the node from  record
			del record[node.data]
		else :
			record[node.data] = record.get(node.data) - 1
		
	
	#  Display inorder sequence in binary tree
	def printInorder(self, node) :
		if (node == None) :
			return
		
		#  Visit left subtree
		self.printInorder(node.left)
		#  print node value
		print("  ", node.data, end = "")
		#  Visit right subtree
		self.printInorder(node.right)
	
	def palindromePath(self) :
		# Use to count frequency of path elements
		record = dict()
		self.result = 0
		self.getPath(self.root, record)
		#  Display inorder of binary tree
		self.printInorder(self.root)
		#  Display calculated result
		print("\n Permutable palindromic paths is ", self.result)
	

def main() :
	#  Create new binary tree
	tree1 = BinaryTree()
	tree2 = BinaryTree()
	#     a   
	#    / \                         
	#   /   \    
	#  c     e    
	# / \     \               
	# a   a     a
	#   / \   / \
	#  d   c e   b
	# -----------------
	# Constructing binary tree   
	tree1.root = TreeNode('a')
	tree1.root.left = TreeNode('c')
	tree1.root.left.right = TreeNode('a')
	tree1.root.left.right.left = TreeNode('d')
	tree1.root.left.right.right = TreeNode('c')
	tree1.root.left.left = TreeNode('a')
	tree1.root.right = TreeNode('e')
	tree1.root.right.right = TreeNode('a')
	tree1.root.right.right.right = TreeNode('b')
	tree1.root.right.right.left = TreeNode('e')
	#  Test A
	#  a  c  a
	#  a  c  c a  or c  a  a  c
	#  a  e  e a  or e  a  a  e
	#  Result : 3
	tree1.palindromePath()
	#     a  
	#    / \                          
	#   /   \    
	#  a     h    
	#   \     \               
	#    a     e
	#   / \   / \
	#  d   a b   h
	#             \
	#              e
	# -----------------
	# Constructing binary tree
	tree2.root = TreeNode('a')
	tree2.root.left = TreeNode('a')
	tree2.root.left.right = TreeNode('a')
	tree2.root.left.right.left = TreeNode('d')
	tree2.root.left.right.right = TreeNode('a')
	tree2.root.right = TreeNode('h')
	tree2.root.right.right = TreeNode('e')
	tree2.root.right.right.right = TreeNode('h')
	tree2.root.right.right.left = TreeNode('b')
	tree2.root.right.right.right.right = TreeNode('e')
	#  Test B
	#  [a  a  a  a]
	#  [h  e  a  e  h] or [e h a h e]
	#  Result : 2
	tree2.palindromePath()

if __name__ == "__main__": main()

input

   a   c   d   a   c   a   e   e   a   b
 Permutable palindromic paths is  3
   a   d   a   a   a   h   b   e   h   e
 Permutable palindromic paths is  2
#  Ruby program
#  Count tree paths whose permutation is palindrome

#  Binary Tree node
class TreeNode 
	# Define the accessor and reader of class TreeNode
	attr_reader :data, :left, :right
	attr_accessor :data, :left, :right
	def initialize(data) 
		#  Set node value
		self.data = data
		self.left = nil
		self.right = nil
	end

end

class BinaryTree 
	# Define the accessor and reader of class BinaryTree
	attr_reader :root, :result
	attr_accessor :root, :result
	def initialize() 
		self.root = nil
		self.result = 0
	end

	def isPalindrome(record) 
		odd = false
		#  Check that if given string can be palindrome
		record.each { | key, value |
			if (value % 2 != 0) 
				if (odd == false) 
					#  One odd allow
					odd = true
				else
					#  When have more than two odd character
					return false
				end

			end

		}
		return true
	end

	def getPath(node, record) 
		if (node == nil) 
			return
		end

		if (record.key?(node.data)) 
			#  Increase node frequency
			record[node.data] = record[node.data] + 1
		else
 
			#  Add new node in record
			record[node.data] = 1
		end

		#  Visit left subtree
		self.getPath(node.left, record)
		#  Visit right subtree
		self.getPath(node.right, record)
		if (node.left == nil && node.right == nil) 
			#  When node is leaf node
			if (self.isPalindrome(record)) 
				#  Increase the resultant counter
				#  When path are generated a palindrome
				self.result += 1
			end

		end

		if (record[node.data] == 1) 
			#  Remove the node from  record
			record.delete(node.data)
		else
 
			record[node.data] = record[node.data] - 1
		end

	end

	#  Display inorder sequence in binary tree
	def printInorder(node) 
		if (node == nil) 
			return
		end

		#  Visit left subtree
		self.printInorder(node.left)
		#  print node value
		print("  ", node.data)
		#  Visit right subtree
		self.printInorder(node.right)
	end

	def palindromePath() 
		# Use to count frequency of path elements
		record = Hash.new()
		self.result = 0
		self.getPath(self.root, record)
		#  Display inorder of binary tree
		self.printInorder(self.root)
		#  Display calculated result
		print("\n Permutable palindromic paths is ", self.result, "\n")
	end

end

def main() 
	#  Create new binary tree
	tree1 = BinaryTree.new()
	tree2 = BinaryTree.new()
	#     a   
	#    / \                         
	#   /   \    
	#  c     e    
	# / \     \               
	# a   a     a
	#   / \   / \
	#  d   c e   b
	# -----------------
	# Constructing binary tree   
	tree1.root = TreeNode.new('a')
	tree1.root.left = TreeNode.new('c')
	tree1.root.left.right = TreeNode.new('a')
	tree1.root.left.right.left = TreeNode.new('d')
	tree1.root.left.right.right = TreeNode.new('c')
	tree1.root.left.left = TreeNode.new('a')
	tree1.root.right = TreeNode.new('e')
	tree1.root.right.right = TreeNode.new('a')
	tree1.root.right.right.right = TreeNode.new('b')
	tree1.root.right.right.left = TreeNode.new('e')
	#  Test A
	#  a  c  a
	#  a  c  c a  or c  a  a  c
	#  a  e  e a  or e  a  a  e
	#  Result : 3
	tree1.palindromePath()
	#     a  
	#    / \                          
	#   /   \    
	#  a     h    
	#   \     \               
	#    a     e
	#   / \   / \
	#  d   a b   h
	#             \
	#              e
	# -----------------
	# Constructing binary tree
	tree2.root = TreeNode.new('a')
	tree2.root.left = TreeNode.new('a')
	tree2.root.left.right = TreeNode.new('a')
	tree2.root.left.right.left = TreeNode.new('d')
	tree2.root.left.right.right = TreeNode.new('a')
	tree2.root.right = TreeNode.new('h')
	tree2.root.right.right = TreeNode.new('e')
	tree2.root.right.right.right = TreeNode.new('h')
	tree2.root.right.right.left = TreeNode.new('b')
	tree2.root.right.right.right.right = TreeNode.new('e')
	#  Test B
	#  [a  a  a  a]
	#  [h  e  a  e  h] or [e h a h e]
	#  Result : 2
	tree2.palindromePath()
end

main()

input

  a  c  d  a  c  a  e  e  a  b
 Permutable palindromic paths is 3
  a  d  a  a  a  h  b  e  h  e
 Permutable palindromic paths is 2
import scala.collection.mutable._;
/*
  Scala program
  Count tree paths whose permutation is palindrome
*/
// Binary Tree node
class TreeNode(var data: Char,
	var left: TreeNode,
		var right: TreeNode)
{
	def this(data: Char)
	{
		// Set node value
		this(data,null,null);
	}
}
class BinaryTree(var root: TreeNode,
	var result: Int)
{
	def this()
	{
		this( null, 0);
	}
	def isPalindrome(record: HashMap[Character, Int]): Boolean = {
		var odd: Boolean = false;
		// Check that if given string can be palindrome
		for ( (key,value) <-  record)
		{
			if (value % 2 != 0)
			{
				if (odd == false)
				{
					// One odd allow
					odd = true;
				}
				else
				{
					// When have more than two odd character
					return false;
				}
			}
		}
		return true;
	}
	def getPath(node: TreeNode, record: HashMap[Character, Int]): Unit = {
		if (node == null)
		{
			return;
		}
		if (record.contains(node.data))
		{
			// Increase node frequency
			record.addOne(node.data, record.get(node.data).get + 1);
		}
		else
		{
			// Add new node in record
			record.addOne(node.data, 1);
		}
		// Visit left subtree
		getPath(node.left, record);
		// Visit right subtree
		getPath(node.right, record);
		if (node.left == null && node.right == null)
		{
			// When node is leaf node
			if (isPalindrome(record))
			{
				// Increase the resultant counter
				// When path are generated a palindrome
				this.result += 1;
			}
		}
		if (record.get(node.data).get == 1)
		{
			// Remove the node from  record
			record.remove(node.data);
		}
		else
		{
			record.addOne(node.data, record.get(node.data).get - 1);
		}
	}
	// Display inorder sequence in binary tree
	def printInorder(node: TreeNode): Unit = {
		if (node == null)
		{
			return;
		}
		// Visit left subtree
		printInorder(node.left);
		// print node value
		print("  " + node.data);
		// Visit right subtree
		printInorder(node.right);
	}
	def palindromePath(): Unit = {
		//Use to count frequency of path elements
		var record: HashMap[Character, Int] = new HashMap[Character, Int]();
		this.result = 0;
		getPath(this.root, record);
		// Display inorder of binary tree
		printInorder(this.root);
		// Display calculated result
		println("\n Permutable palindromic paths is " + this.result);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		// Create new binary tree
		var tree1: BinaryTree = new BinaryTree();
		var tree2: BinaryTree = new BinaryTree();
		/*
		     a   
		    / \                         
		   /   \    
		  c     e    
		 / \     \               
		a   a     a
		   / \   / \
		  d   c e   b
		-----------------
		Constructing binary tree   
		*/
		tree1.root = new TreeNode('a');
		tree1.root.left = new TreeNode('c');
		tree1.root.left.right = new TreeNode('a');
		tree1.root.left.right.left = new TreeNode('d');
		tree1.root.left.right.right = new TreeNode('c');
		tree1.root.left.left = new TreeNode('a');
		tree1.root.right = new TreeNode('e');
		tree1.root.right.right = new TreeNode('a');
		tree1.root.right.right.right = new TreeNode('b');
		tree1.root.right.right.left = new TreeNode('e');
		// Test A
		// a  c  a
		// a  c  c a  or c  a  a  c
		// a  e  e a  or e  a  a  e
		// Result : 3
		tree1.palindromePath();
		/*
		     a  
		    / \                          
		   /   \    
		  a     h    
		   \     \               
		    a     e
		   / \   / \
		  d   a b   h
		             \
		              e
		-----------------
		Constructing binary tree
		*/
		tree2.root = new TreeNode('a');
		tree2.root.left = new TreeNode('a');
		tree2.root.left.right = new TreeNode('a');
		tree2.root.left.right.left = new TreeNode('d');
		tree2.root.left.right.right = new TreeNode('a');
		tree2.root.right = new TreeNode('h');
		tree2.root.right.right = new TreeNode('e');
		tree2.root.right.right.right = new TreeNode('h');
		tree2.root.right.right.left = new TreeNode('b');
		tree2.root.right.right.right.right = new TreeNode('e');
		// Test B
		// [a  a  a  a]
		// [h  e  a  e  h] or [e h a h e]
		// Result : 2
		tree2.palindromePath();
	}
}

input

  a  c  d  a  c  a  e  e  a  b
 Permutable palindromic paths is 3
  a  d  a  a  a  h  b  e  h  e
 Permutable palindromic paths is 2
import Foundation;
/*
  Swift 4 program
  Count tree paths whose permutation is palindrome
*/

// Binary Tree node
class TreeNode
{
	var data: Character;
	var left: TreeNode? ;
	var right: TreeNode? ;
	init(_ data: Character)
	{
		// Set node value
		self.data = data;
		self.left = nil;
		self.right = nil;
	}
}
class BinaryTree
{
	var root: TreeNode? ;
	var result: Int;
	init()
	{
		self.root = nil;
		self.result = 0;
	}
	func isPalindrome(_ record: [Character : Int]) -> Bool
	{
		var odd = false;
		// Check that if given string can be palindrome
		for (_, value) in record
		{
			if (value % 2  != 0)
			{
				if (odd == false)
				{
					// One odd allow
					odd = true;
				}
				else
				{
					// When have more than two odd character
					return false;
				}
			}
		}
		return true;
	}
	func getPath(_ node: TreeNode? , _ record : inout[Character : Int])
	{
		if (node == nil)
		{
			return;
		}
		if (record.keys.contains(node!.data))
		{
			// Increase node frequency
			record[node!.data] = record[node!.data]! + 1;
		}
		else
		{
			// Add new node in record
			record[node!.data] = 1;
		}
		// Visit left subtree
		self.getPath(node!.left, &record);
		// Visit right subtree
		self.getPath(node!.right, &record);
		if (node!.left == nil && node!.right == nil)
		{
			// When node is leaf node
			if (self.isPalindrome(record))
			{
				// Increase the resultant counter
				// When path are generated a palindrome
				self.result += 1;
			}
		}
		if (record[node!.data] == 1)
		{
			// Remove the node from  record
			record.removeValue(forKey: node!.data);
		}
		else
		{
			record[node!.data] = record[node!.data]! - 1;
		}
	}
	// Display inorder sequence in binary tree
	func printInorder(_ node: TreeNode? )
	{
		if (node == nil)
		{
			return;
		}
		// Visit left subtree
		self.printInorder(node!.left);
		// print node value
		print("  ", node!.data, terminator: "");
		// Visit right subtree
		self.printInorder(node!.right);
	}
	func palindromePath()
	{
		//Use to count frequency of path elements
		var record = [Character : Int]();
		self.result = 0;
		self.getPath(self.root, &record);
		// Display inorder of binary tree
		self.printInorder(self.root);
		// Display calculated result
		print("\n Permutable palindromic paths is ", self.result);
	}
}
func main()
{
	// Create new binary tree
	let tree1 = BinaryTree();
	let tree2 = BinaryTree();
	/*
	     a   
	    / \                         
	   /   \    
	  c     e    
	 / \     \               
	a   a     a
	   / \   / \
	  d   c e   b
	-----------------
	Constructing binary tree   
	*/
	tree1.root = TreeNode("a");
	tree1.root!.left = TreeNode("c");
	tree1.root!.left!.right = TreeNode("a");
	tree1.root!.left!.right!.left = TreeNode("d");
	tree1.root!.left!.right!.right = TreeNode("c");
	tree1.root!.left!.left = TreeNode("a");
	tree1.root!.right = TreeNode("e");
	tree1.root!.right!.right = TreeNode("a");
	tree1.root!.right!.right!.right = TreeNode("b");
	tree1.root!.right!.right!.left = TreeNode("e");
	// Test A
	// a  c  a
	// a  c  c a  or c  a  a  c
	// a  e  e a  or e  a  a  e
	// Result : 3
	tree1.palindromePath();
	/*
	     a  
	    / \                          
	   /   \    
	  a     h    
	   \     \               
	    a     e
	   / \   / \
	  d   a b   h
	             \
	              e
	-----------------
	Constructing binary tree
	*/
	tree2.root = TreeNode("a");
	tree2.root!.left = TreeNode("a");
	tree2.root!.left!.right = TreeNode("a");
	tree2.root!.left!.right!.left = TreeNode("d");
	tree2.root!.left!.right!.right = TreeNode("a");
	tree2.root!.right = TreeNode("h");
	tree2.root!.right!.right = TreeNode("e");
	tree2.root!.right!.right!.right = TreeNode("h");
	tree2.root!.right!.right!.left = TreeNode("b");
	tree2.root!.right!.right!.right!.right = TreeNode("e");
	// Test B
	// [a  a  a  a]
	// [h  e  a  e  h] or [e h a h e]
	// Result : 2
	tree2.palindromePath();
}
main();

input

   a   c   d   a   c   a   e   e   a   b
 Permutable palindromic paths is  3
   a   d   a   a   a   h   b   e   h   e
 Permutable palindromic paths is  2
/*
  Kotlin program
  Count tree paths whose permutation is palindrome
*/
// Binary Tree node
class TreeNode
{
	var data: Char;
	var left: TreeNode ? ;
	var right: TreeNode ? ;
	constructor(data: Char)
	{
		// Set node value
		this.data = data;
		this.left = null;
		this.right = null;
	}
}
class BinaryTree
{
	var root: TreeNode ? ;
	var result: Int;
	constructor()
	{
		this.root = null;
		this.result = 0;
	}
	fun isPalindrome(record: MutableMap < Char, Int > ): Boolean
	{
		var odd: Boolean = false;
		// Check that if given string can be palindrome
		for ((_, value) in record)
		{
			if (value % 2 != 0)
			{
				if (odd == false)
				{
					// One odd allow
					odd = true;
				}
				else
				{
					// When have more than two odd character
					return false;
				}
			}
		}
		return true;
	}
	fun getPath(node: TreeNode ? , record : MutableMap < Char, Int >): Unit
	{
		if (node == null)
		{
			return;
		}
		if (record.containsKey(node.data))
		{
			// Increase node frequency
			record.put(node.data, record.getValue(node.data) + 1);
		}
		else
		{
			// Add new node in record
			record.put(node.data, 1);
		}
		// Visit left subtree
		this.getPath(node.left, record);
		// Visit right subtree
		this.getPath(node.right, record);
		if (node.left == null && node.right == null)
		{
			// When node is leaf node
			if (this.isPalindrome(record))
			{
				// Increase the resultant counter
				// When path are generated a palindrome
				this.result += 1;
			}
		}
		if (record.getValue(node.data) == 1)
		{
			// Remove the node from  record
			record.remove(node.data);
		}
		else
		{
			record.put(node.data, record.getValue(node.data) - 1);
		}
	}
	// Display inorder sequence in binary tree
	fun printInorder(node: TreeNode ? ): Unit
	{
		if (node == null)
		{
			return;
		}
		// Visit left subtree
		this.printInorder(node.left);
		// print node value
		print("  " + node.data);
		// Visit right subtree
		this.printInorder(node.right);
	}
	fun palindromePath(): Unit
	{
		// Use to count frequency of path elements
		val record = mutableMapOf < Char , Int > ();
		this.result = 0;
		this.getPath(this.root, record);
		// Display inorder of binary tree
		this.printInorder(this.root);
		// Display calculated result
		println("\n Permutable palindromic paths is " + this.result);
	}
}
fun main(args: Array < String > ): Unit
{
	// Create new binary tree
	val tree1: BinaryTree = BinaryTree();
	val tree2: BinaryTree = BinaryTree();
	/*
	     a   
	    / \                         
	   /   \    
	  c     e    
	 / \     \               
	a   a     a
	   / \   / \
	  d   c e   b
	-----------------
	Constructing binary tree   
	*/
	tree1.root = TreeNode('a');
	tree1.root?.left = TreeNode('c');
	tree1.root?.left?.right = TreeNode('a');
	tree1.root?.left?.right?.left = TreeNode('d');
	tree1.root?.left?.right?.right = TreeNode('c');
	tree1.root?.left?.left = TreeNode('a');
	tree1.root?.right = TreeNode('e');
	tree1.root?.right?.right = TreeNode('a');
	tree1.root?.right?.right?.right = TreeNode('b');
	tree1.root?.right?.right?.left = TreeNode('e');
	// Test A
	// a  c  a
	// a  c  c a  or c  a  a  c
	// a  e  e a  or e  a  a  e
	// Result : 3
	tree1.palindromePath();
	/*
	     a  
	    / \                          
	   /   \    
	  a     h    
	   \     \               
	    a     e
	   / \   / \
	  d   a b   h
	             \
	              e
	-----------------
	Constructing binary tree
	*/
	tree2.root = TreeNode('a');
	tree2.root?.left = TreeNode('a');
	tree2.root?.left?.right = TreeNode('a');
	tree2.root?.left?.right?.left = TreeNode('d');
	tree2.root?.left?.right?.right = TreeNode('a');
	tree2.root?.right = TreeNode('h');
	tree2.root?.right?.right = TreeNode('e');
	tree2.root?.right?.right?.right = TreeNode('h');
	tree2.root?.right?.right?.left = TreeNode('b');
	tree2.root?.right?.right?.right?.right = TreeNode('e');
	// Test B
	// [a  a  a  a]
	// [h  e  a  e  h] or [e h a h e]
	// Result : 2
	tree2.palindromePath();
}

input

  a  c  d  a  c  a  e  e  a  b
 Permutable palindromic paths is 3
  a  d  a  a  a  h  b  e  h  e
 Permutable palindromic paths is 2


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