> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/trees/pseudo-palindromic-paths-in-a-binary-tree.md).

# 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:**

![](https://assets.leetcode.com/uploads/2020/05/06/palindromic_paths_1.png)

```
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:**

![](https://assets.leetcode.com/uploads/2020/05/07/palindromic_paths_2.png)

```
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`.

```java
class Solution {
    int ans = 0;
    int[] freq;

    public int pseudoPalindromicPaths(TreeNode root) {
        if (root == null)
            return 0;
        freq = new int[10];
        helper(root, 0);
        return ans;
    }

    public void helper(TreeNode root, int oddCount) {
        if (root.left == null && root.right == null) {
            // For a plaindrome permutation ,we must have atmost 1 odd frequency character
            freq[root.val]++;
            if (freq[root.val] % 2 != 0)
                oddCount++;
            else
                oddCount--;
            if (oddCount <= 1)
                ans++;
            freq[root.val]--;
            return;
        }
        freq[root.val]++;
        if (freq[root.val] % 2 != 0)
            oddCount++;
        else
            oddCount--;
        if (root.left != null)
            helper(root.left, oddCount);
        if (root.right != null)
            helper(root.right, oddCount);
        freq[root.val]--;
        return;
    }
}
```