# 1457. Pseudo-Palindromic Paths in a Binary Tree

#### Medium

***

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)

<pre><code>Input: root = [2,3,1,3,1,null,1]
<strong>Output:
</strong> 2 
<strong>Explanation:
</strong> 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).
</code></pre>

**Example 2:**

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

<pre><code>Input: root = [2,1,1,1,3,null,null,null,null,null,1]
<strong>Output:
</strong> 1 
<strong>Explanation:
</strong> 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).
</code></pre>

**Example 3:**

<pre><code>Input: root = [9]
<strong>Output:
</strong> 1
</code></pre>

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**Constraints:**

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

```python
class Solution:
    def pseudoPalindromicPaths (self, root: Optional[TreeNode]) -> int:
        return self.dfs(root)
    
    def dfs(self, root, s = set()):
        if not root:
            return 0
        s.remove(root.val) if root.val in s else s.add(root.val)
        count = 0
        if not root.left and not root.right:
            # If set is empty then it is even plaindrome -> no swap required
            # If set has 1 element, it is odd plaindrome -> count it
            # if set has > 1 element, then it is not plaindrome -> reject
            if len(s) <= 1:
                count = 1
        else:
            count += self.dfs(root.left, s) + self.dfs(root.right, s)
        # Again removing since element may be added in subtree
        s.remove(root.val) if root.val in s else s.add(root.val)
        return count
```