Maximum number in Binary tree of binary values

Last Updated : 18 Jan, 2024

Given a binary tree consisting of nodes, each containing a binary value of either 0 or 1, the task is to find the maximum decimal number that can be formed by traversing from the root to a leaf node. The maximum number is achieved by concatenating the binary values along the path from the root to a leaf.

Examples:

Input:

1
/ \
0 1
/ \ / \
0 1 1 0

Output: 7
Explaination: The possible binary numbers are 100, 101, 111, 110, So the maximum number in decimal is 7 (111).

Input:

1
/ \
1 0
/ \ \
0 1 0
/
1

Output: 15
Explaination: The possible binary numbers are 110, 1111, 100. So the maximum number in decimal is 15 (1111).

Approach: To solve the problem follow the below Intuition:

The algorithm in binary tree is a DFS recursive traversal, accumulating the binary number as it traverses the tree from the root to the leaves. At each node, the binary number is updated by shifting it left (equivalent to multiplying by 2) and adding the binary value of the current node. The algorithm considers both left and right subtrees, comparing their maximum binary numbers. By the end of the traversal, the algorithm returns the maximum binary number that can be formed from the root to any leaf in the binary tree.

Follow the steps to solve the problem:

Declare a currentNumber, which is the binary number formed from the root node to the current node.
Travese recusively and if the current node is NULL (i.e., a leaf node is reached, thus base case hits), then return 0 because there is no number to add to the currentNumber.
The currentNumber is updated by shifting it left (equivalent to multiplying by 2) and adding the binary value of the current node (0 or 1) using bitwise OR operation.
The function then recursively calls itself for the left and right children, passing the updated currentNumber. It calculates the maximum binary number for the left and right subtrees.
Finally, returns the maximum of the currentNumber and the maximum binary numbers obtained from the left and right subtrees.

Below is the implementation for the above approach:

C++14

// C++ implementation of above approach #include <bits/stdc++.h> using namespace std;  // Bulid tree class class Node { public:     int data;     Node* left;     Node* right;     Node(int val)     {         data = val;         left = NULL;         right = NULL;     } };  int findMaxBinaryNumber(Node* root, int currentNumber) {     if (root == NULL) {         return 0;     }      // Update the currentNumber by shifting     // left and adding the node's value     currentNumber = (currentNumber << 1) | root->data;      // Recursively find the maximum number     // in the left and right subtrees     int leftMax         = findMaxBinaryNumber(root->left, currentNumber);     int rightMax         = findMaxBinaryNumber(root->right, currentNumber);      // Return the maximum of the current subtree     return max(currentNumber, max(leftMax, rightMax)); }  // Driver Code int main() {      // Create a binary tree     Node* root = new Node(1);     root->left = new Node(0);     root->right = new Node(1);     root->left->left = new Node(0);     root->left->right = new Node(1);     root->right->left = new Node(1);     root->right->right = new Node(0);      // Find the maximum binary number in the tree     int maxNumber = findMaxBinaryNumber(root, 0);     cout << "Maximum Binary Number: " << maxNumber;      return 0; }

Java

// Java implementation of the above approach import java.util.*;  // Build tree class class Node {     int data;     Node left, right;      public Node(int val)     {         data = val;         left = right = null;     } }  public class GFG {     // Function to find the maximum binary number     static int findMaxBinaryNumber(Node root,                                    int currentNumber)     {         if (root == null) {             return 0;         }          // Update the currentNumber by shifting         // left and adding the node's value         currentNumber = (currentNumber << 1) | root.data;          // Recursively find the maximum number         // in the left and right subtrees         int leftMax             = findMaxBinaryNumber(root.left, currentNumber);         int rightMax = findMaxBinaryNumber(root.right,                                            currentNumber);          // Return the maximum of the current subtree         return Math.max(currentNumber,                         Math.max(leftMax, rightMax));     }      // Driver code     public static void main(String[] args)     {         // Create a binary tree         Node root = new Node(1);         root.left = new Node(0);         root.right = new Node(1);         root.left.left = new Node(0);         root.left.right = new Node(1);         root.right.left = new Node(1);         root.right.right = new Node(0);          // Find the maximum binary number in the tree         int maxNumber = findMaxBinaryNumber(root, 0);         System.out.println("Maximum Binary Number: "                            + maxNumber);     } }  // This code is contributed by Susobhan Akhuli

Python3

class Node:     def __init__(self, val):         self.data = val         self.left = None         self.right = None  def find_max_binary_number(root, current_number):     if not root:         return 0      # Update the current_number by shifting left and adding the node's value     current_number = (current_number << 1) | root.data      # Recursively find the maximum number in the left and right subtrees     left_max = find_max_binary_number(root.left, current_number)     right_max = find_max_binary_number(root.right, current_number)      # Return the maximum of the current subtree     return max(current_number, max(left_max, right_max))  # Driver Code if __name__ == "__main__":     # Create a binary tree     root = Node(1)     root.left = Node(0)     root.right = Node(1)     root.left.left = Node(0)     root.left.right = Node(1)     root.right.left = Node(1)     root.right.right = Node(0)      # Find the maximum binary number in the tree     max_number = find_max_binary_number(root, 0)     print("Maximum Binary Number:", max_number)  # This code is cotributed by akshitaguprzj3

using System;  // Build tree class class Node {     public int data;     public Node left;     public Node right;      public Node(int val)     {         data = val;         left = null;         right = null;     } }  class GFG {     static int FindMaxBinaryNumber(Node root, int currentNumber)     {         if (root == null)         {             return 0;         }          // Update the currentNumber by shifting         // left and adding the node's value         currentNumber = (currentNumber << 1) | root.data;          // Recursively find the maximum number         // in the left and right subtrees         int leftMax = FindMaxBinaryNumber(root.left, currentNumber);         int rightMax = FindMaxBinaryNumber(root.right, currentNumber);          // Return the maximum of the current subtree         return Math.Max(currentNumber, Math.Max(leftMax, rightMax));     }      // Driver Code     static void Main()     {         // Create a binary tree         Node root = new Node(1);         root.left = new Node(0);         root.right = new Node(1);         root.left.left = new Node(0);         root.left.right = new Node(1);         root.right.left = new Node(1);         root.right.right = new Node(0);          // Find the maximum binary number in the tree         int maxNumber = FindMaxBinaryNumber(root, 0);         Console.WriteLine("Maximum Binary Number: " + maxNumber);     } }

JavaScript

// JavaScript implementation of above approach  // Build tree class class Node {     constructor(val) {         this.data = val;         this.left = null;         this.right = null;     } }  function findMaxBinaryNumber(root, currentNumber) {     if (root === null) {         return 0;     }      // Update the currentNumber by shifting     // left and adding the node's value     currentNumber = (currentNumber << 1) | root.data;      // Recursively find the maximum number     // in the left and right subtrees     let leftMax = findMaxBinaryNumber(root.left, currentNumber);     let rightMax = findMaxBinaryNumber(root.right, currentNumber);      // Return the maximum of the current subtree     return Math.max(currentNumber, Math.max(leftMax, rightMax)); }  // Create a binary tree let root = new Node(1); root.left = new Node(0); root.right = new Node(1); root.left.left = new Node(0); root.left.right = new Node(1); root.right.left = new Node(1); root.right.right = new Node(0);  // Find the maximum binary number in the tree let maxNumber = findMaxBinaryNumber(root, 0); console.log("Maximum Binary Number: " + maxNumber);