题目：

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.

思路：

题目给了hint，用DFS记录整条path上的数，然后出现次数为奇数的数字只能小于等于1，否则不对称，这个逻辑很清楚就不证明了。这里我选择的容器时哈希set，每次遇到一个数先检查有没有，如果没有就加入set，如果有就说明已经出现过一次了，从set中删去，这样最后走完的时候，set中剩余的size就是这条路上出现次数为奇数的数的数量(有点绕不过逻辑很简单)。这里还有一个点就是需要检查叶子节点，dfs的base case不能是当前root为空，因为叶子节点会造成左右两个空结点，这样最终算出来的结果是真实结果的两倍，这里可能会想到直接把结果除以2即可，这也是错的。因为有可能某个节点是有左无右，这样就凭白造成了一个空结点，也就是一个结算点。因此要确认到叶子节点再做成base case，这就导致在初始函数我们要先确认给定root是不是为空。

代码：

/**
 * 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) {}
 * };
 */
class Solution {
public:
    int pseudoPalindromicPaths (TreeNode* root) {
        if(!root)
            return 0;
        unordered_set<int> s;
        check(root,s);
        return count;
    }
private:
    int count=0;
    void check(TreeNode* root, unordered_set<int> s)
    {
        if(s.count(root->val))
            s.erase(root->val);
        else
            s.insert(root->val);
        if(!root->left&&!root->right)
        {
            if(s.size()<=1)
                count+=1;
            return ;
        }
        if(root->left)
            check(root->left,s);
        if(root->right)
            check(root->right,s);
    }
};