Count of 1’s in any path in a Binary Tree

Given a binary tree of 0s and 1s, the task is to find the maximum number of 1s in any path in the tree. The path may start and end at any node in the tree.

Example:

Input: 
       1
      / \
     0   1
    / \
   1   1
      / \
     1   0

Output: 4

Approach:



  1. A function countUntil has been created which returns the maximum count of 1 in any vertical path below that node.
  2. This path must be a single path and this path must include at-most one child of the node as well as its itself. i.e countUntil returns the max count of 1 in either left child or right child and also includes itself in count if its value is 1.
  3. So, from any node countUntil(node->left) + countUntil(node->right) + node->value will give the number of 1s in the path which contains the node and its left and right path and no ancestor of the node is considered.
  4. Taking the maximum of all nodes will give the required answer.

Below is the implementation of the above approach

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// A binary tree node
struct Node {
    int data;
    struct Node *left, *right;
};
  
// A utility function to allocate a new node
struct Node* newNode(int data)
{
    struct Node* newNode = new Node;
    newNode->data = data;
    newNode->left = newNode->right = NULL;
    return (newNode);
}
  
// This function updates overall count of 1 in 'res'
// And returns count 1s going through root.
int countUntil(Node* root, int& res)
{
    // Base Case
    if (root == NULL)
        return 0;
  
    // l and r store count of 1s going through left and
    // right child of root respectively
    int l = countUntil(root->left, res);
    int r = countUntil(root->right, res);
  
    // maxCount represents the count of 1s when the Node under
    // consideration is the root of the maxCount path and no
    // ancestors of the root are there in maxCount path
    int maxCount;
  
    // if the value at node is 1 then its
    // count will be considered
    // including the leftCount and the rightCount
    if (root->data == 1)
        maxCount = l + r + 1;
    else
        maxCount = l + r;
  
    // Store the Maximum Result.
    res = max(res, maxCount);
  
    // return max count in a single path.
    // This path must include at-most one child
    // of the root as well as itself
  
    // if the value at node is 1
    // then its count will be considered
    // including the maximum of leftCount or the rightCount
    if (root->data == 1)
        return max(l, r) + 1;
    else
        return max(l, r);
}
  
// Returns maximum count of 1 in any path
// in tree with given root
int findMaxCount(Node* root)
{
    // Initialize result
    int res = INT_MIN;
  
    // Compute and return result
    countUntil(root, res);
    return res;
}
  
// Driver program
int main(void)
{
    struct Node* root = newNode(1);
    root->left = newNode(0);
    root->right = newNode(1);
    root->left->left = newNode(1);
    root->left->right = newNode(1);
    root->left->right->left = newNode(1);
    root->left->right->right = newNode(0);
    cout << findMaxCount(root);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the above approach
class GFG
{
  
    // A binary tree node
    static class Node 
    {
        int data;
        Node left, right;
    };
  
    static int res;
  
    // A utility function to allocate a new node
    static Node newNode(int data)
    {
        Node newNode = new Node();
        newNode.data = data;
        newNode.left = newNode.right = null;
        return (newNode);
    }
  
    // This function updates overall count of 1 in 'res'
    // And returns count 1s going through root.
    static int countUntil(Node root) 
    {
        // Base Case
        if (root == null)
            return 0;
  
        // l and r store count of 1s going through left and
        // right child of root respectively
        int l = countUntil(root.left);
        int r = countUntil(root.right);
  
        // maxCount represents the count of 1s when the Node under
        // consideration is the root of the maxCount path and no
        // ancestors of the root are there in maxCount path
        int maxCount;
  
        // if the value at node is 1 then its
        // count will be considered
        // including the leftCount and the rightCount
        if (root.data == 1)
            maxCount = l + r + 1;
        else
            maxCount = l + r;
  
        // Store the Maximum Result.
        res = Math.max(res, maxCount);
  
        // return max count in a single path.
        // This path must include at-most one child
        // of the root as well as itself
  
        // if the value at node is 1
        // then its count will be considered
        // including the maximum of leftCount or the rightCount
        if (root.data == 1)
            return Math.max(l, r) + 1;
        else
            return Math.max(l, r);
    }
  
    // Returns maximum count of 1 in any path
    // in tree with given root
    static int findMaxCount(Node root) 
    {
        // Initialize result
        res = Integer.MIN_VALUE;
  
        // Compute and return result
        countUntil(root);
        return res;
    }
  
    // Driver program
    public static void main(String[] args)
    {
        Node root = newNode(1);
        root.left = newNode(0);
        root.right = newNode(1);
        root.left.left = newNode(1);
        root.left.right = newNode(1);
        root.left.right.left = newNode(1);
        root.left.right.right = newNode(0);
        System.out.print(findMaxCount(root));
    }
}
  
// This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

      
// C# implementation of the above approach
using System;
  
class GFG
{
  
    // A binary tree node
    class Node 
    {
        public int data;
        public Node left, right;
    };
  
    static int res;
  
    // A utility function to allocate a new node
    static Node newNode(int data)
    {
        Node newNode = new Node();
        newNode.data = data;
        newNode.left = newNode.right = null;
        return (newNode);
    }
  
    // This function updates overall count of 1 in 'res'
    // And returns count 1s going through root.
    static int countUntil(Node root) 
    {
        // Base Case
        if (root == null)
            return 0;
  
        // l and r store count of 1s going through left and
        // right child of root respectively
        int l = countUntil(root.left);
        int r = countUntil(root.right);
  
        // maxCount represents the count of 1s when the Node under
        // consideration is the root of the maxCount path and no
        // ancestors of the root are there in maxCount path
        int maxCount;
  
        // if the value at node is 1 then its
        // count will be considered
        // including the leftCount and the rightCount
        if (root.data == 1)
            maxCount = l + r + 1;
        else
            maxCount = l + r;
  
        // Store the Maximum Result.
        res = Math.Max(res, maxCount);
  
        // return max count in a single path.
        // This path must include at-most one child
        // of the root as well as itself
  
        // if the value at node is 1
        // then its count will be considered
        // including the maximum of leftCount or the rightCount
        if (root.data == 1)
            return Math.Max(l, r) + 1;
        else
            return Math.Max(l, r);
    }
  
    // Returns maximum count of 1 in any path
    // in tree with given root
    static int findMaxCount(Node root) 
    {
        // Initialize result
        res = int.MinValue;
  
        // Compute and return result
        countUntil(root);
        return res;
    }
  
    // Driver program
    public static void Main(String[] args)
    {
        Node root = newNode(1);
        root.left = newNode(0);
        root.right = newNode(1);
        root.left.left = newNode(1);
        root.left.right = newNode(1);
        root.left.right.left = newNode(1);
        root.left.right.right = newNode(0);
        Console.Write(findMaxCount(root));
    }
}
  
// This code is contributed by PrinciRaj1992

chevron_right


Output:

4

Time Complexity: O(n)
where n is number of nodes in Binary Tree.



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : 29AjayKumar, princiraj1992