Introduction

In this blog post, we will explore how to solve the LeetCode problem "1457" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution:

• High-Level Approach:

  • Perform a Depth-First Search (DFS) traversal of the binary tree.
  • Maintain a frequency count of digits (1-9) encountered along each path from the root to a leaf.
  • For each leaf node, check if the path is pseudo-palindromic by counting the number of digits with odd frequencies. A path is pseudo-palindromic if at most one digit has an odd frequency.

• Complexity:

  • Runtime Complexity: O(N), where N is the number of nodes in the tree.
  • Storage Complexity: O(H), where H is the height of the tree, due to the recursion stack. In the worst case (skewed tree), H can be N.

Code

    class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def pseudoPalindromicPaths(root: TreeNode) -> int:
    """
    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.
    """

    def dfs(node, counts):
        if not node:
            return 0

        counts[node.val] += 1
        if not node.left and not node.right:
            odd_count = 0
            for i in range(1, 10):
                if counts[i] % 2 != 0:
                    odd_count += 1
            counts[node.val] -= 1  # Backtrack! Decrement count before returning
            return 1 if odd_count <= 1 else 0

        left_paths = dfs(node.left, counts) if node.left else 0
        right_paths = dfs(node.right, counts) if node.right else 0

        counts[node.val] -= 1  # Backtrack! Decrement count before returning
        return left_paths + right_paths

    counts = [0] * 10  # counts[i] stores the count of digit i (1 to 9)
    return dfs(root, counts)