Length of longest straight path from a given Binary Tree

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Given a Binary Tree, the task is to find the length of the longest straight path of the given binary tree.

Straight Path is defined as the path that starts from any node and ends at another node in the tree such that the direction of traversal from the source node to the destination node always remains the same i.e., either left or right, without any change in direction that is left->left ->left or right->right->right direction.

Examples:

Input:

Output: 2
Explanation:
The path shown in green is the longest straight path from 4 to 6 which is of length 2.

Input:

Output: 3
Explanation:
The path shown in green is the longest straight path from 5 to 0 which is of length 3.

Approach: The idea is to use the Postorder traversal. Follow the steps below to solve the problem:

For each node, check the direction of the current Node (either left or right) and check which direction of its child is providing the longest length below it to that node.
If the direction of the current node and the child giving the longest length is not the same then save the result of that child and pass the length of the other child to its parent.
Using the above steps find the longest straight path at each node and save the result to print the maximum value among all the straight paths.
Print the maximum path after the above steps.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a Tree node
struct Node {
    int key;
    struct Node *left, *right;
};
 
// Function to create a new node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}
 
// Function to find the longest
// straight path in a tree
int findPath(Node* root, char name,
             int& max_v)
{
    // Base Case
    if (root == NULL) {
        return 0;
    }
 
    // Recursive call on left child
    int left = findPath(root->left,
                        'l', max_v);
 
    // Recursive call on right child
    int right = findPath(root->right,
                         'r', max_v);
 
    // Return the maximum straight
    // path possible from current node
    if (name == 't') {
        return max(left, right);
    }
 
    // Leaf node
    if (left == 0 && right == 0) {
        return 1;
    }
 
    // Executes when either of the
    // child is present or both
    else {
 
        // Pass the longest value from
        // either direction
        if (left < right) {
            if (name == 'r')
                return 1 + right;
 
            else {
                max_v = max(max_v, right);
                return 1 + left;
            }
        }
        else {
            if (name == 'l')
                return 1 + left;
 
            else {
                max_v = max(max_v, left);
                return 1 + right;
            }
        }
    }
    return 0;
}
 
// Driver Code
int main()
{
 
    // Given Tree
    Node* root = newNode(3);
 
    root->left = newNode(3);
    root->right = newNode(3);
 
    root->left->right = newNode(2);
    root->right->left = newNode(4);
    root->right->left->left = newNode(4);
 
    int max_v = max(
        findPath(root, 't', max_v),
        max_v);
 
    // Print the maximum length
    cout << max_v << "\n";
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
static int max_v;
 
// Structure of a Tree node
static class Node
{
    int key;
    Node left, right;
};
 
// Function to create a new node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}
 
// Function to find the longest
// straight path in a tree
static int findPath(Node root, char name)
{
     
    // Base Case
    if (root == null)
    {
        return 0;
    }
 
    // Recursive call on left child
    int left = findPath(root.left, 'l');
 
    // Recursive call on right child
    int right = findPath(root.right, 'r');
 
    // Return the maximum straight
    // path possible from current node
    if (name == 't')
    {
        return Math.max(left, right);
    }
 
    // Leaf node
    if (left == 0 && right == 0)
    {
        return 1;
    }
 
    // Executes when either of the
    // child is present or both
    else
    {
         
        // Pass the longest value from
        // either direction
        if (left < right) 
        {
            if (name == 'r')
                return 1 + right;
            else
            {
                max_v = Math.max(max_v, right);
                return 1 + left;
            }
        }
        else
        {
            if (name == 'l')
                return 1 + left;
            else
            {
                max_v = Math.max(max_v, left);
                return 1 + right;
            }
        }
    }
}
 
// Driver Code
public static void main(String[] args)
{
 
    // Given Tree
    Node root = newNode(3);
 
    root.left = newNode(3);
    root.right = newNode(3);
    root.left.right = newNode(2);
    root.right.left = newNode(4);
    root.right.left.left = newNode(4);
 
    max_v = Math.max(findPath(root, 't'),
                     max_v);
 
    // Print the maximum length
    System.out.print(max_v+ "\n");
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
max_v = 0
 
# Structure of a Tree node
class newNode:
     
    def __init__(self, key):
         
        self.key = key
        self.left = None
        self.right = None
 
# Function to find the longest
# straight path in a tree
def findPath(root, name):
     
    global max_v
     
    # Base Case
    if (root == None):
        return 0
 
    # Recursive call on left child
    left = findPath(root.left, 'l')
 
    # Recursive call on right child
    right = findPath(root.right, 'r')
     
    # Return the maximum straight
    # path possible from current node
    if (name == 't'):
        return max(left, right)
 
    # Leaf node
    if (left == 0 and right == 0):
        return 1
 
    # Executes when either of the
    # child is present or both
    else:
         
        # Pass the longest value from
        # either direction
        if (left < right):
            if (name == 'r'):
                return 1 + right
            else:
                max_v = max(max_v, right)
                return 1 + left
        else:
            if (name == 'l'):
                return 1 + left
            else:
                max_v = max(max_v, left)
                return 1 + right
                 
    return 0
 
def helper(root):
     
    global max_v
    temp = max(findPath(root, 't'), max_v)
    print(temp)
 
# Driver Code
if __name__ == '__main__':
     
    # Given Tree
    root = newNode(3)
    root.left = newNode(3)
    root.right = newNode(3)
    root.left.right = newNode(2)
    root.right.left = newNode(4)
    root.right.left.left = newNode(4)
     
    helper(root)
 
# This code is contributed by ipg2016107

C#

// C# program for 
// the above approach
using System;
class GFG{
     
static int max_v;
 
// Structure of a Tree node
public class Node
{
  public int key;
  public Node left, right;
};
 
// Function to create a new node
static Node newNode(int key)
{
  Node temp = new Node();
  temp.key = key;
  temp.left = temp.right = null;
  return (temp);
}
 
// Function to find the longest
// straight path in a tree
static int findPath(Node root, 
                    char name)
{
  // Base Case
  if (root == null)
  {
    return 0;
  }
 
  // Recursive call on left child
  int left = findPath(root.left, 'l');
 
  // Recursive call on right child
  int right = findPath(root.right, 'r');
 
  // Return the maximum straight
  // path possible from current node
  if (name == 't')
  {
    return Math.Max(left, right);
  }
 
  // Leaf node
  if (left == 0 && right == 0)
  {
    return 1;
  }
 
  // Executes when either of the
  // child is present or both
  else
  {
    // Pass the longest value from
    // either direction
    if (left < right) 
    {
      if (name == 'r')
        return 1 + right;
      else
      {
        max_v = Math.Max(max_v, right);
        return 1 + left;
      }
    }
    else
    {
      if (name == 'l')
        return 1 + left;
      else
      {
        max_v = Math.Max(max_v, left);
        return 1 + right;
      }
    }
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given Tree
  Node root = newNode(3);
 
  root.left = newNode(3);
  root.right = newNode(3);
  root.left.right = newNode(2);
  root.right.left = newNode(4);
  root.right.left.left = newNode(4);
 
  max_v = Math.Max(findPath(root, 't'), max_v);
 
  // Print the maximum length
  Console.Write(max_v + "\n");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

       // JavaScript code for the above approach
       // Structure of a Tree node
       class Node {
           constructor(key) {
               this.key = key;
               this.left = null;
               this.right = null;
           }
       }
 
       // Function to create a new node
       function newNode(key) {
           const temp = new Node(key);
           return temp;
       }
 
       // Function to find the longest
       // straight path in a tree
       function findPath(root, name, maxV) {
           // Base Case
           if (root === null) {
               return 0;
           }
 
           // Recursive call on left child
           const left = findPath(root.left, 'l', maxV);
 
           // Recursive call on right child
           const right = findPath(root.right, 'r', maxV);
 
           // Return the maximum straight
           // path possible from current node
           if (name === 't') {
               return Math.max(left, right);
           }
 
           // Leaf node
           if (left === 0 && right === 0) {
               return 1;
           }
 
           // Executes when either of the
           // child is present or both
           else {
               // Pass the longest value from
               // either direction
               if (left < right) {
                   if (name === 'r') return 1 + right;
                   else {
                       maxV[0] = Math.max(maxV[0], right);
                       return 1 + left;
                   }
               } else {
                   if (name === 'l') return 1 + left;
                   else {
                       maxV[0] = Math.max(maxV[0], left);
                       return 1 + right;
                   }
               }
           }
           return 0;
       }
 
       // Driver Code
 
       // Given Tree
       const root = newNode(3);
 
       root.left = newNode(3);
       root.right = newNode(3);
 
       root.left.right = newNode(2);
       root.right.left = newNode(4);
       root.right.left.left = newNode(4);
 
       const maxV = [0];
       const maxVal = Math.max(
           findPath(root, 't', maxV),
           maxV[0]
       );
 
       // Print the maximum length
       console.log(maxVal);
 
// This code is contributed by Potta Lokesh.

Output:

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

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