29. Pseudo-Palindromic Paths in a Binary Tree.cpp

/*
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 given binary tree will have between 1 and 10^5 nodes.
Node values are digits from 1 to 9.

Hint #1  
Note that the node values of a path form a palindrome if at most one digit has an odd frequency (parity).

Hint #2  
Use a Depth First Search (DFS) keeping the frequency (parity) of the digits. Once you are in a leaf node check if at most one digit has an odd frequency (parity).
*/

/**
 * 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
{
  bool palendrome(vector<int> &arr)
  {
    int one = 0;
    for (int i = 1; i < 10; ++i)
    {
      if (arr[i] % 2 != 0)
      {
        if (one == 0)
          one = 1;
        else
          return false;
      }
    }
    return true;
  }

  int dfs(TreeNode *root, vector<int> arr)
  {
    if (!root)
      return 0;
    if (!root->left && !root->right)
    {
      arr[root->val]++;
      if (palendrome(arr))
        return 1;
      return 0;
    }

    int ans = 0;
    arr[root->val]++;

    if (root->left)
      ans += dfs(root->left, arr);
    if (root->right)
      ans += dfs(root->right, arr);

    return ans;
  }

public:
  int pseudoPalindromicPaths(TreeNode *root)
  {
    vector<int> arr(10, 0);
    return dfs(root, arr);
  }
};