Pseudo-Palindromic Paths in a Binary Tree

• Initialize a global or member variable count to 0.
• Create a frequency array counts of size 10, initialized to all zeros, to store the frequency of digits 1 through 9.
• Define a recursive DFS function, let's call it dfs(node, counts).
• In the dfs function:
  • If the current node is null, return immediately.
  • Increment the frequency of the current node's value: counts[node.val]++.
  • Check if the current node is a leaf node (i.e., node.left == null && node.right == null).
    • If it is a leaf, check if the current path is pseudo-palindromic. This is done by a helper function that iterates through the counts array and counts how many digits have an odd frequency. If the number of odd-frequency digits is 0 or 1, increment the global count.
  • Recursively call the function for the left and right children: dfs(node.left, counts) and dfs(node.right, counts).
  • Backtrack: After the recursive calls for the children return, decrement the frequency of the current node's value: counts[node.val]--. This step is crucial to correctly reflect the path state as the traversal unwinds.
• Start the traversal by calling dfs(root, counts) from the main function.
• Return the final count.

A path's node values can form a palindrome if at most one digit appears an odd number of times. We can implement a recursive DFS function, say dfs(node, counts), that explores the tree. The counts parameter is an array that stores the frequency of each digit (1-9) from the root to the current node.

The main function initializes a counter for pseudo-palindromic paths to zero and an empty frequency array, then calls dfs(root, counts). The key to this approach is backtracking. After visiting a node and its descendants, we must undo the change made to the frequency count for that node. This ensures that when we explore a sibling branch, the counts accurately represent the path to that sibling's parent, not the path through the previously explored cousin nodes.

1/**2 * Definition for a binary tree node.3 * public class TreeNode {4 *     int val;5 *     TreeNode left;6 *     TreeNode right;7 *     TreeNode() {}8 *     TreeNode(int val) { this.val = val; }9 *     TreeNode(int val, TreeNode left, TreeNode right) {10 *         this.val = val;11 *         this.left = left;12 *         this.right = right;13 *     }14 * }15 */16class Solution {17    int count = 0;18 19    public int pseudoPalindromicPaths (TreeNode root) {20        int[] pathCounts = new int[10];21        dfs(root, pathCounts);22        return count;23    }24 25    private void dfs(TreeNode node, int[] pathCounts) {26        if (node == null) {27            return;28        }29 30        // Add current node to the path31        pathCounts[node.val]++;32 33        // If it's a leaf node, check for pseudo-palindrome34        if (node.left == null && node.right == null) {35            if (isPseudoPalindrome(pathCounts)) {36                count++;37            }38        } else {39            // Continue traversal40            dfs(node.left, pathCounts);41            dfs(node.right, pathCounts);42        }43 44        // Backtrack: remove current node from the path45        pathCounts[node.val]--;46    }47 48    private boolean isPseudoPalindrome(int[] counts) {49        int oddCount = 0;50        for (int i = 1; i <= 9; i++) {51            if (counts[i] % 2 != 0) {52                oddCount++;53            }54        }55        return oddCount <= 1;56    }57}