LeetCode 1457: Pseudo-Palindromic Paths in a Binary Tree Solution

Master LeetCode problem 1457 (Pseudo-Palindromic Paths in a Binary Tree), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1457. Pseudo-Palindromic Paths in a Binary Tree

Medium

Bit Manipulation Tree Depth-First Search Breadth-First Search Binary Tree

Problem Explanation

Explanation

To solve this problem, we can perform a depth-first search (DFS) on the binary tree while keeping track of the frequency of each digit encountered in the path from the root to the current node. At each node, we check if the path ending at that node forms a pseudo-palindromic path. We can do this by checking if there is at most one digit with an odd frequency in the path.

Algorithm:

Initialize a count variable to keep track of the number of pseudo-palindromic paths.
Implement a recursive function that performs a DFS traversal of the binary tree and keeps track of the frequency of each digit in the current path.
At each node, update the frequency of the current node's value in the path.
Check if the current path forms a pseudo-palindromic path by counting the number of digits with odd frequencies. Increment the count variable if at most one digit has an odd frequency.
Recursively call the function on the left and right child nodes.
Return the count of pseudo-palindromic paths.

Time Complexity:

The time complexity of this approach is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once during the DFS traversal.

Space Complexity:

The space complexity of this approach is O(H), where H is the height of the binary tree. This is the space used for the recursive call stack during the DFS traversal.

Solution Code

class Solution {
    public int pseudoPalindromicPaths (TreeNode root) {
        return dfs(root, 0);
    }
    
    private int dfs(TreeNode node, int path) {
        if (node == null) return 0;
        
        path ^= (1 << node.val);
        
        int count = 0;
        if (node.left == null && node.right == null) {
            if ((path & (path - 1)) == 0) {
                count = 1;
            }
        }
        
        count += dfs(node.left, path) + dfs(node.right, path);
        
        return count;
    }
}

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