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Longest path to the bottom of a Binary Tree forming an Arithmetic Progression

  • Difficulty Level : Medium
  • Last Updated : 01 Jul, 2021

Given a Binary Tree consisting of N nodes, the task is to find the length of the longest path from any node to the bottom of the tree such that all the node values form an Arithmetic Progression.

Examples:

Input:

Output: 4
Explanation:
From the above given tree, the longest path with node values forming an AP is {6, 9, 12, 15} (Length = 4).



Input:

Output: 2

Approach: The given problem can be solved by using recursion and performing the DFS Traversal on the given tree. The idea is to keep track of the difference between the current root node and the next descendant node and update the length of the longest path accordingly. Follow the steps below to solve the given problem:

  • Initialize a variable, say maxLength as 1 that stores the maximum length of the path from any node to the bottom of the tree forming an Arithmetic Progression.
  • Define a function, say dfs(root, currentDifference, count, maxLength) that takes the current root node, current difference, count of nodes forming AP, and the resultant maximum length as the parameter and perform the following steps:
    • If the root’s left node exists, then perform the following steps:
      • Find the difference between the value of the root and its left node.
      • If the difference is found to be currentDifference, then update the value of maxLength to the maximum of maxLength and (count + 1) and recursively call for the function dfs(root->left, currentDifference, count + 1, maxLength).
      • Otherwise, recursively call for the dfs(root->left, difference, 2, maxLength).
    • If the root’s right node exists, then perform the following steps:
      • Find the difference between the value of the root and its right node.
      • If the difference is found to be currentDifference, then update the value of maxLength to the maximum of maxLength and (count + 1) and recursively call for the function dfs(root->right, currentDifference, count + 1, maxLength).
      • Otherwise, recursively call for the dfs(root->left, difference, 2, maxLength).
  • If the left child of the given root node exists, then call for the dfs(root->left, difference, 2, maxLength) where the difference is the difference between the value of the root and its left node.
  • If the right child of the given root node exists, then call for the dfs(root->right, difference, 2, maxLength) where the difference is the difference between the value of the root and its right node.
  • After completing the above steps, print the value of maxLength as the resultant maximum length of the path from any node to the bottom of the tree forming an Arithmetic Progression.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include "bits/stdc++.h"
using namespace std;
 
// Structure of the Tree Node
struct Tree {
    int val;
    Tree *left, *right;
};
 
// Function to create a new node
Tree* newNode(int data)
{
    Tree* temp = new Tree();
    temp->val = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Function to perform DFS Traversal
// to find the maximum length of a path
// to a bottom node forming an AP
void dfs(Tree* root, int currentDifference,
         int count, int& maxLength)
{
    // If the root's left child exists
    if (root->left) {
 
        // Calculate the difference
        int difference = root->left->val
                         - root->val;
 
        // If the difference is same
        // as the current difference
        if (difference == currentDifference) {
            dfs(root->left, currentDifference,
                count + 1, maxLength);
 
            // Update the maxLength
            maxLength = max(maxLength,
                            count + 1);
        }
 
        // Otherwise
        else {
            dfs(root->left, difference,
                2, maxLength);
        }
    }
 
    // If the root's right child exists
    if (root->right) {
 
        // Find the difference
        int difference = root->right->val
                         - root->val;
 
        // If the difference is the same
        // as the current difference
        if (difference == currentDifference) {
 
            dfs(root->right, currentDifference,
                count + 1, maxLength);
 
            // Update the maxLength
            maxLength = max(maxLength,
                            count + 1);
        }
 
        // Otherwise
        else {
            dfs(root->right, difference,
                2, maxLength);
        }
    }
}
 
// Function to find the maximum length
// of the path from any node to bottom
// of the tree forming an AP
int maximumLengthAP(Tree* root)
{
 
    // Base Cases
    if (root == NULL)
        return 0;
 
    if (root->left == NULL
        and root->right == NULL) {
        return 1;
    }
 
    // Stores the resultant
    // maximum length of the path
    int maxLength = 2;
 
    // If the root's left child exists
    if (root->left) {
 
        int difference = root->left->val
                         - root->val;
        dfs(root->left, difference, 2,
            maxLength);
    }
 
    // If the root's right child exists
    if (root->right) {
        int difference = root->right->val
                         - root->val;
        dfs(root->right, difference, 2,
            maxLength);
    }
 
    // Return the maximum length obtained
    return maxLength;
}
 
// Driver Code
int main()
{
 
    // Given Tree
    Tree* root = newNode(6);
    root->right = newNode(9);
    root->right->left = newNode(7);
    root->right->right = newNode(12);
    root->right->right->right = newNode(15);
 
    cout << maximumLengthAP(root);
 
    return 0;
}

Java




// Java program for the above approach
import java.lang.*;
import java.util.*;
 
class GFG{
     
static int maxLength;   
     
// TreeNode class
static class Node
{
    public int val;
    public Node left, right;
};
 
static Node newNode(int key)
{
    Node temp = new Node();
    temp.val = key;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to perform DFS Traversal
// to find the maximum length of a path
// to a bottom node forming an AP
static void dfs(Node root, int currentDifference,
                int count)
{
     
    // If the root's left child exists
    if (root.left != null)
    {
         
        // Calculate the difference
        int difference = root.left.val - root.val;
 
        // If the difference is same
        // as the current difference
        if (difference == currentDifference)
        {
            dfs(root.left, currentDifference,
                count + 1);
 
            // Update the maxLength
            maxLength = Math.max(maxLength,
                                 count + 1);
        }
 
        // Otherwise
        else
        {
            dfs(root.left, difference, 2);
        }
    }
 
    // If the root's right child exists
    if (root.right != null)
    {
         
        // Find the difference
        int difference = root.right.val - root.val;
 
        // If the difference is the same
        // as the current difference
        if (difference == currentDifference)
        {
            dfs(root.right, currentDifference,
                count + 1);
 
            // Update the maxLength
            maxLength = Math.max(maxLength,
                                 count + 1);
        }
 
        // Otherwise
        else
        {
            dfs(root.right, difference, 2);
        }
    }
}
 
// Function to find the maximum length
// of the path from any node to bottom
// of the tree forming an AP
static int maximumLengthAP(Node root)
{
     
    // Base Cases
    if (root == null)
        return 0;
 
    if (root.left == null &&
        root.right == null)
    {
        return 1;
    }
 
    // Stores the resultant
    // maximum length of the path
     maxLength = 2;
 
    // If the root's left child exists
    if (root.left != null)
    {
        int difference = root.left.val - root.val;
        dfs(root.left, difference, 2);
    }
 
    // If the root's right child exists
    if (root.right != null)
    {
        int difference = root.right.val - root.val;
        dfs(root.right, difference, 2);
    }
 
    // Return the maximum length obtained
    return maxLength;
}   
 
// Driver code
public static void main(String[] args)
{
 
    // Given Tree
    Node root = newNode(6);
    root.right = newNode(9);
    root.right.left = newNode(7);
    root.right.right = newNode(12);
    root.right.right.right = newNode(15);
     
    System.out.println(maximumLengthAP(root));
}
}
 
// This code is contributed by offbeat

Javascript




<script>
 
    // JavaScript program for the above approach
     
    let maxLength;  
     
    class Node
    {
        constructor(key) {
           this.left = null;
           this.right = null;
           this.val = key;
        }
    }
     
    function newNode(key)
    {
        let temp = new Node(key);
        return temp;
    }
 
    // Function to perform DFS Traversal
    // to find the maximum length of a path
    // to a bottom node forming an AP
    function dfs(root, currentDifference, count)
    {
 
        // If the root's left child exists
        if (root.left != null)
        {
 
            // Calculate the difference
            let difference = root.left.val - root.val;
 
            // If the difference is same
            // as the current difference
            if (difference == currentDifference)
            {
                dfs(root.left, currentDifference,
                    count + 1);
 
                // Update the maxLength
                maxLength = Math.max(maxLength,
                                     count + 1);
            }
 
            // Otherwise
            else
            {
                dfs(root.left, difference, 2);
            }
        }
 
        // If the root's right child exists
        if (root.right != null)
        {
 
            // Find the difference
            let difference = root.right.val - root.val;
 
            // If the difference is the same
            // as the current difference
            if (difference == currentDifference)
            {
                dfs(root.right, currentDifference,
                    count + 1);
 
                // Update the maxLength
                maxLength = Math.max(maxLength,
                                     count + 1);
            }
 
            // Otherwise
            else
            {
                dfs(root.right, difference, 2);
            }
        }
    }
 
    // Function to find the maximum length
    // of the path from any node to bottom
    // of the tree forming an AP
    function maximumLengthAP(root)
    {
 
        // Base Cases
        if (root == null)
            return 0;
 
        if (root.left == null &&
            root.right == null)
        {
            return 1;
        }
 
        // Stores the resultant
        // maximum length of the path
         maxLength = 2;
 
        // If the root's left child exists
        if (root.left != null)
        {
            let difference = root.left.val - root.val;
            dfs(root.left, difference, 2);
        }
 
        // If the root's right child exists
        if (root.right != null)
        {
            let difference = root.right.val - root.val;
            dfs(root.right, difference, 2);
        }
 
        // Return the maximum length obtained
        return maxLength;
    
     
    // Given Tree
    let root = newNode(6);
    root.right = newNode(9);
    root.right.left = newNode(7);
    root.right.right = newNode(12);
    root.right.right.right = newNode(15);
      
    document.write(maximumLengthAP(root));
     
</script>
Output: 
4

 

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

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