【leetcode】1457. Pseudo-Palindromic Paths in a Binary Tree

题目如下：

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.

解题思路：要构成伪回文串，需要满足一个条件，出现次数为奇数的数字最多只能有一个。所以，只需要在遍历树的过程中，记录遍历路径每个数字出现的次数即可。如果一个数字出现了偶数次，其实和出现0次是没有区别的。因此，我用二进制来保存每个数字出现的次数。

代码如下：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    res = 0
    def pseudoPalindromicPaths (self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def recursive(node,num):
            num ^= pow(2,node.val - 1)
            if node.left == None and node.right == None:
                bs = bin(num)[2:]
                if bs.count('1') <= 1:
                    self.res += 1
                return
            if node.left != None:
                recursive(node.left,num)
            if node.right != None:
                recursive(node.right,num)

        self.res = 0
        recursive(root,0)
        return self.res