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

<p><em>Return the number of <strong>pseudo-palindromic</strong> paths going from the root node to leaf nodes.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/05/06/palindromic_paths_1.png" style="width: 300px; height: 201px;" /></p>

<pre>
<strong>Input:</strong> 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).
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/05/07/palindromic_paths_2.png" style="width: 300px; height: 314px;" /></strong></p>

<pre>
<strong>Input:</strong> 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).
</pre>

<p><strong class="example">Example 3:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 9</code></li>
</ul>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pseudoPalindromicPaths (self, root: Optional[TreeNode]) -> int:
        def can_form_palindrome(num_to_freq: dict):
            used_odd = False
            for num, freq in num_to_freq.items():
                if freq % 2 == 1:
                    if used_odd == False:
                        used_odd = True
                    else:
                        print(used_odd)
                        return False
            return True
        res = 0
        num_to_freq = {}
        def dfs(node):
            nonlocal res
            if node == None:
                return
            num_to_freq[node.val] = num_to_freq.get(node.val, 0) + 1
            if node.left == None and node.right == None:    # determine end of path
                print(num_to_freq)
                if can_form_palindrome(num_to_freq):
                    res += 1
            dfs(node.left)
            if node.left != None:
                num_to_freq[node.left.val] -= 1
                if num_to_freq[node.left.val] == 0:
                    num_to_freq.pop(node.left.val)
            dfs(node.right)
            if node.right != None:
                num_to_freq[node.right.val] -= 1
                if num_to_freq[node.right.val] == 0:
                    num_to_freq.pop(node.right.val)
        dfs(root)
        return res
43500