Maximum value of Bitwise AND from root to leaf in a Binary tree

Given a Binary Tree, the task is to find the maximum value of Bitwise AND from any path from the root node to the leaf node.

Examples:

Input: Below is the given graph:

Output: 7
Explanation:
path 1: 15->3->5 = (15 & 3 & 5) = 1
path 2: 15->3->1 =(15 & 3 & 1) = 1
path 3: 15->7->31=(15 & 7 & 31)= 7 (maximum)
path 4: 15->7->9 = (15 & 7 & 9) =1, out of these 7 is the maximum.

Input: Below is the given graph:

Output: 6
Explanation:
Path 1: 31->3->7  =  (31 & 3 & 7)  = 3
Path 2: 31->3->1 =  (31 & 3 & 1)  = 1
Path 3: 31->15->5 =  (31 & 15 & 5)  5
Path 4: 31->15->22 =  (31 & 15 & 22)  = 6, out of these 6 is the maximum.

Approach: The idea is to traverse all the path from the root node to the leaf node and calculate the Bitwise AND of all the nodes that occurred in that path. Keep a global variable to update the maximum Bitwise AND value from all the paths.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Initialise to update the maximum
// AND value from all the path
int maxm = 0;

// Node structure
struct Node {
    int val;

    // Left & right child of the node
    Node *left, *right;

    // Initialize constructor
    Node(int x)
    {
        val = x;
        left = NULL;
        right = NULL;
    }
};

// Function to find the maximum value
// of Bitwise AND from root to leaf
// in a Binary tree
void maxm_Anding(Node* root, int ans)
{
    // Check if root is not null
    if (!root)
        return;

    if (root->left == NULL
        and root->right == NULL) {
        ans &= root->val;

        // Find the maximum AND value and
        // store in global maxm variable
        maxm = max(ans, maxm);

        return;
    }

    // Traverse left of binary tree
    maxm_Anding(root->left,
                ans & root->val);

    // Traverse right of the binary tree
    maxm_Anding(root->right,
                ans & root->val);
}

// Driver Code
int main()
{
    // Given Tree
    Node* root = new Node(15);
    root->left = new Node(3);
    root->right = new Node(7);
    root->left->left = new Node(5);
    root->left->right = new Node(1);
    root->right->left = new Node(31);
    root->right->right = new Node(9);

    // Function Call
    maxm_Anding(root, root->val);

    // Print the maximum AND value
    cout << maxm << endl;

    return 0;
}

Java

// Java program for the above approach
class GFG{

// Initialise to update the maximum
// AND value from all the path
static int maxm = 0;

// Node structure
static class Node
{
    int val;

    // Left & right child of the node
    Node left, right;

    // Initialize constructor
    Node(int x)
    {
        val = x;
        left = null;
        right = null;
    }
};

// Function to find the maximum value
// of Bitwise AND from root to leaf
// in a Binary tree
static void maxm_Anding(Node root, int ans)
{
    // Check if root is not null
    if (root == null)
        return;

    if (root.left == null && root.right == null)
    {
        ans &= root.val;

        // Find the maximum AND value and
        // store in global maxm variable
        maxm = Math.max(ans, maxm);

        return;
    }

    // Traverse left of binary tree
    maxm_Anding(root.left,
                ans & root.val);

    // Traverse right of the binary tree
    maxm_Anding(root.right,
                ans & root.val);
}

// Driver Code
public static void main(String[] args)
{
    // Given Tree
    Node root = new Node(15);
    root.left = new Node(3);
    root.right = new Node(7);
    root.left.left = new Node(5);
    root.left.right = new Node(1);
    root.right.left = new Node(31);
    root.right.right = new Node(9);

    // Function Call
    maxm_Anding(root, root.val);

    // Print the maximum AND value
    System.out.print(maxm + "\n");
}
}

// This code is contributed by sapnasingh4991

Python3

# Python3 program for the above approach

# Initialise to update the maximum
# AND value from all the path
maxm = 0

# Node structure
class Node:

    def __init__(self, x):

        self.val = x

        # Left & right child of the node
        self.left = None
        self.right = None

# Function to find the maximum value
# of Bitwise AND from root to leaf
# in a Binary tree
def maxm_Anding(root: Node, ans: int) -> None:

    global maxm

    # Check if root is not null
    if not root:
        return

    if (root.left is None and
        root.right is None):
        ans &= root.val

        # Find the maximum AND value and
        # store in global maxm variable
        maxm = max(ans, maxm)

        return

    # Traverse left of binary tree
    maxm_Anding(root.left, ans & root.val)

    # Traverse right of the binary tree
    maxm_Anding(root.right, ans & root.val)

# Driver Code
if __name__ == "__main__":

    # Given Tree
    root = Node(15)
    root.left = Node(3)
    root.right = Node(7)
    root.left.left = Node(5)
    root.left.right = Node(1)
    root.right.left = Node(31)
    root.right.right = Node(9)

    # Function Call
    maxm_Anding(root, root.val)

    # Print the maximum AND value
    print(maxm)

# This code is contributed by sanjeev2552

C#

// C# program for the above approach
using System;
class GFG{

// Initialise to update the maximum
// AND value from all the path
static int maxm = 0;

// Node structure
class Node
{
    public int val;

    // Left & right child of the node
    public Node left, right;

    // Initialize constructor
    public Node(int x)
    {
        val = x;
        left = null;
        right = null;
    }
};

// Function to find the maximum value
// of Bitwise AND from root to leaf
// in a Binary tree
static void maxm_Anding(Node root, int ans)
{
    // Check if root is not null
    if (root == null)
        return;

    if (root.left == null && root.right == null)
    {
        ans &= root.val;

        // Find the maximum AND value and
        // store in global maxm variable
        maxm = Math.Max(ans, maxm);

        return;
    }

    // Traverse left of binary tree
    maxm_Anding(root.left,
                ans & root.val);

    // Traverse right of the binary tree
    maxm_Anding(root.right,
                ans & root.val);
}

// Driver Code
public static void Main(String[] args)
{
    // Given Tree
    Node root = new Node(15);
    root.left = new Node(3);
    root.right = new Node(7);
    root.left.left = new Node(5);
    root.left.right = new Node(1);
    root.right.left = new Node(31);
    root.right.right = new Node(9);

    // Function Call
    maxm_Anding(root, root.val);

    // Print the maximum AND value
    Console.Write(maxm + "\n");
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// JavaScript program for the above approach
let maxm = 0;

// Node structure
class Node
{

    // Initialize constructor
    constructor(x)
    {
        this.val = x;
        this.left = null;
        this.right = null;
    }
}

var root;

// Function to find the maximum value
// of Bitwise AND from root to leaf
// in a Binary tree
function maxm_Anding(root, ans)
{

    // Check if root is not null
    if (!root)
        return;

    if (root.left == null &&
       root.right == null)
    {
        ans &= root.val;

        // Find the maximum AND value and
        // store in global maxm variable
        maxm = Math.max(ans, maxm);

        return;
    }

    // Traverse left of binary tree
    maxm_Anding(root.left,
                ans & root.val);

    // Traverse right of the binary tree
    maxm_Anding(root.right,
                ans & root.val);
}

// Driver Code
root = new Node(15);
root.left = new Node(3);
root.right = new Node(7);
root.left.left = new Node(5);
root.left.right = new Node(1);
root.right.left = new Node(31);
root.right.right = new Node(9);

// Function Call
maxm_Anding(root, root.val);

document.write(maxm);

// This code is contributed by Dharanendra L V.

</script>

Output:

7

Time Complexity: O(N²)
Auxiliary Space: O(1)