1457. Pseudo-Palindromic Paths in a Binary Tree | Leetcode solution

 1457. Pseudo-Palindromic Paths in a Binary Tree

Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be pseudo-palindromic if at least one permutation of the node values in the path is a palindrome.

Return the number of pseudo-palindromic paths going from the root node to leaf nodes.

 

Example 1:

Input: root = [2,3,1,3,1,null,1]
Output: 2 
Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the red path [2,3,3], the green path [2,1,1], and the path [2,3,1]. Among these paths only red path and green path are pseudo-palindromic paths since the red path [2,3,3] can be rearranged in [3,2,3] (palindrome) and the green path [2,1,1] can be rearranged in [1,2,1] (palindrome).

Example 2:

Input: root = [2,1,1,1,3,null,null,null,null,null,1]
Output: 1 
Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the green path [2,1,1], the path [2,1,3,1], and the path [2,1]. Among these paths only the green path is pseudo-palindromic since [2,1,1] can be rearranged in [1,2,1] (palindrome).

Example 3:

Input: root = [9]
Output: 1

 

Constraints:

• The number of nodes in the tree is in the range [1, 105].
• 1 <= Node.val <= 9

solution:

/**

 * Definition for a binary tree node.

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}

 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}

 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

 * };

 */

class Solution {

public:

    void paths(TreeNode * root , vector<int>t ,int &count )

    {

         if(root->left == NULL && root->right == NULL)

         { 

            t[root->val]++;

            int g = 0;

            int flag = 0;

            for(int i = 0 ; i <10 ; i++)

            {    

                if(t[i] %2 != 0)

                {

                    if(flag != 1)

                        flag =1;

                    else

                    {   

                        g =1 ;

                        break;

                    }

                }

            }

            if(g == 0)

               count++;

           return;

         }

         

        if(root->left != NULL)

        {   

            t[root->val]++;

            paths(root->left, t, count );

            t[root->val]--;

        }

        if(root->right != NULL)

        {   

            t[root->val]++;

            paths(root->right, t , count);

            t[root->val]--;

        }

             

    }

    int pseudoPalindromicPaths (TreeNode* root) {

        

         vector<int>t(10, 0);

        int count = 0;

        paths(root , t, count);

        

        

        return count;

        

    }

};

references : https://mycodevillage.blogspot.com/2022/09/root-to-leaf-paths-binary-tree.html