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 number of nodes in the tree is in the range
[1, 105]
. 1 <= Node.val <= 9
Solution
pseudo-palindromic-paths-in-a-binary-tree.py
# 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 dfs(self, root, mp):
if not root: return 0
mp[root.val] += 1
count = 0
if not root.left and not root.right and sum(v&1 for v in mp.values()) <= 1:
count += 1
count += self.dfs(root.left, mp)
count += self.dfs(root.right, mp)
mp[root.val] -= 1
return count
def pseudoPalindromicPaths (self, root: TreeNode) -> int:
return self.dfs(root, collections.defaultdict(int))