Description
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.
分析
题目的意思是：判断二叉树从根结点到叶子结点的数能否构成回文子串。找出能构成回文子串的个数，我的思路也很直接，从根结点遍历到叶子结点之后，然后判断一下是否是回文子串就行了。
代码
# 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 solve(self,root,ans):
if(root is None):
return
ans.append(root.val)
if(root.left is None and root.right is None):
self.res+=self.isPalindrome(ans)
self.solve(root.left,ans)
self.solve(root.right,ans)
ans.pop(-1)
def isPalindrome(self,ans):
counter=collections.Counter(ans)
cnt=0
for k,v in counter.most_common():
if(v%2==1):
cnt+=1
if(cnt>1):
return 0
return 1
def pseudoPalindromicPaths (self, root: TreeNode) -> int:
self.res=0
self.solve(root,[])
return self.res