Pseudo-Palindromic Paths in a Binary Tree

Pseudo-Palindromic Paths in a Binary Tree - Problem

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.

Note: A palindrome reads the same forward and backward. For example, 121 is a palindrome while 123 is not.

Input & Output

Example 1 — Basic Tree

$ Input: root = [2,3,1,3,1,null,1]

› Output: 2

💡 Note: Path [2,3,3] has digits 2(×1), 3(×2) - only one odd count, can form palindrome '323'. Path [2,1,1] has digits 2(×1), 1(×2) - only one odd count, can form palindrome '121'. Path [2,3,1] has all odd counts - cannot form palindrome.

Example 2 — Single Path

$ Input: root = [2,1,1,1,3,null,null,null,null,null,1]

› Output: 1

💡 Note: Only one root-to-leaf path exists: [2,1,1,1]. Digit frequencies: 2(×1), 1(×3). Only one digit has odd count, so palindrome '11211' is possible.

Example 3 — No Palindromes

$ Input: root = [9]

› Output: 1

💡 Note: Single node path [9] has only one digit with count 1 (odd). Single digits are always palindromes.

Constraints

The number of nodes in the tree is in the range [1, 10⁵]
1 ≤ Node.val ≤ 9

Visualization

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Asked in

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The key insight is that a palindrome can have at most one character with odd frequency. We can use bit manipulation with XOR to efficiently track odd/even counts as we traverse root-to-leaf paths. Best approach uses DFS with bitmask: Time O(n), Space O(h).

Common Approaches

✓ Bit Manipulation Optimization

⏱️ Time: O(n) Space: O(h)

Instead of using a frequency map, use a single integer as bitmask where each bit represents whether a digit appears odd number of times. XOR operations toggle bits efficiently.

DFS with Frequency Count

⏱️ Time: O(n) Space: O(h + k)

Traverse all root-to-leaf paths using DFS, collect node values in each path, then check if any permutation can form a palindrome by counting digit frequencies.

Bit Manipulation Optimization — Algorithm Steps

Use bitmask where bit i represents odd count of digit i
XOR current digit's bit when visiting/leaving nodes
At leaf nodes, check if bitmask has at most 1 bit set

Visualization

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Step-by-Step Walkthrough

XOR Toggle

XOR digit bit when visiting nodes

Bitmask State

Each bit represents odd count of corresponding digit

Palindrome Check

At most 1 bit set means palindrome possible

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
};

struct TreeNode* createNode(int val) {
    struct TreeNode* node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    node->val = val;
    node->left = NULL;
    node->right = NULL;
    return node;
}

static struct TreeNode* queue[10000];
static int values[10000];
static int is_null[10000];

struct TreeNode* buildTree(int* vals, int* nulls, int len) {
    if (len == 0 || nulls[0]) {
        return NULL;
    }
    
    struct TreeNode* root = createNode(vals[0]);
    queue[0] = root;
    int front = 0, rear = 1;
    int i = 1;
    
    while (front < rear && i < len) {
        struct TreeNode* node = queue[front++];
        
        if (i < len && !nulls[i]) {
            node->left = createNode(vals[i]);
            queue[rear++] = node->left;
        }
        i++;
        
        if (i < len && !nulls[i]) {
            node->right = createNode(vals[i]);
            queue[rear++] = node->right;
        }
        i++;
    }
    
    return root;
}

int dfs(struct TreeNode* node, int bitmask) {
    if (!node) {
        return 0;
    }
    
    bitmask ^= (1 << node->val);
    
    int result = 0;
    
    if (!node->left && !node->right) {
        if (bitmask == 0 || (bitmask & (bitmask - 1)) == 0) {
            result = 1;
        }
    } else {
        result = dfs(node->left, bitmask) + dfs(node->right, bitmask);
    }
    
    return result;
}

int solution(struct TreeNode* root) {
    if (!root) {
        return 0;
    }
    return dfs(root, 0);
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    int len = 0;
    char* ptr = line;
    
    // Skip initial '['
    while (*ptr && *ptr != '[') ptr++;
    if (*ptr) ptr++;
    
    while (*ptr && *ptr != ']') {
        // Skip whitespace and commas
        while (*ptr && (*ptr == ' ' || *ptr == ',')) ptr++;
        
        if (*ptr == ']') break;
        
        if (strncmp(ptr, "null", 4) == 0) {
            is_null[len] = 1;
            values[len] = 0;
            ptr += 4;
        } else if (isdigit(*ptr) || *ptr == '-') {
            is_null[len] = 0;
            values[len] = strtol(ptr, &ptr, 10);
        } else {
            break;
        }
        len++;
    }
    
    struct TreeNode* root = buildTree(values, is_null, len);
    int result = solution(root);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Visit each node once in DFS traversal

✓ Linear Growth

Space Complexity

O(h)

Only recursion stack depth, no extra data structures

✓ Linear Space

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Related Problems

Common Approaches

Bit Manipulation Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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