<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>

// Idea: Tree Traversal. just maintain the OddSet of numbers with odd-counts.
class Solution {
public:
    int pseudoPalindromicPaths(TreeNode* root) {
        int palins = 0; // count of pseudo palindromes
        unordered_set<int> oddSet;
        traverse(root, oddSet, palins);
        return palins;
    }

private:
    static void traverse(TreeNode* p, unordered_set<int>& oddSet, int& palins) {
        if (!p) {
            return;
        }
        // toggle the val in the oddSet
        if (oddSet.count(p->val)) {
            oddSet.erase(p->val);
        } else {
            oddSet.insert(p->val);
        }
        if (!p->left && !p->right) { // is leaf
            if (oddSet.size() <= 1) {
                // good pseudo palindrome
                ++palins;
            }
        } else {
            traverse(p->left, oddSet, palins);
            traverse(p->right, oddSet, palins);
        }

        // restore the oddSet
        if (oddSet.count(p->val)) {
            oddSet.erase(p->val);
        } else {
            oddSet.insert(p->val);
        }
    }
};
/**
 * 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) {}
 * };
 */
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