[leetcode] 1457. Pseudo-Palindromic Paths in a Binary Tree

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.

analysis

The meaning of the question is: to determine whether the number of the binary tree from the root node to the leaf node can form a palindrome substring. To find out the number of palindrome substrings, my idea is also very straightforward. After traversing from the root node to the leaf nodes, I then judge whether it is a palindrome substring.

Code

# 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

references

[LeetCode] [Java/C++/Python] At most one odd occurrence