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.

```
class Solution {
private int ans = 0;
public int pseudoPalindromicPaths (TreeNode root) {
int[] freq = new int[10];
helper(root, freq);
return ans;
}
private void helper(TreeNode root, int[] freq){
if(root== null){
return ;
}
freq[root.val]++;
if(root.left == null && root.right == null){
// this is a leaf node
if(isPalindromicPermutation(freq)){
ans++;
}
}
helper(root.left, freq);
helper(root.right, freq);
freq[root.val]--;
}
private boolean isPalindromicPermutation(int[] freq){
boolean oddFreqFound = false;
for(int el: freq){
if(el%2!=0){
if(oddFreqFound){
return false;
} else{
oddFreqFound = true;
}
}
}
return true;
}
}
```

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