Python|Dfs回溯解二叉树伪回文

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: TreeNode):
         self.ass=[0,0,0,0,0,0,0,0,0,0]         #计数路径中数字的个数
        self.res=0              #记录符合路径的个数
        self.dfs(root,[])     #深搜回溯
        return self.res
    def dfs(self,root,path):    #回溯方法
        if root is None:
            return
        self.ass[root.val]+=1    #将节点数值对应的数字个数加1
        if root.left is None and  root.right is None:#当达到叶子节点时开始判断是否为伪回文
            x=0
            for i in range(10):
                if self.ass[i]%2!=0 and  self.ass[i]!=0:#记录数字个数为奇数的数目
                    x+=1
            if x<2:
                self.res+=1#如果路径中数字个数为奇数的数目为1或0，就是伪回文
        if root.left:
            self.dfs(root.left,path)
        if root.right:
            self.dfs(root.right,path)
        self.ass[root.val]-=1