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++ 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 = 0;
max_v = max(findPath(root, 't' , max_v), max_v);
// Print the maximum length
cout << max_v << "\n" ;
} |
// 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 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# 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 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. |
2
Time Complexity: O(N)
Auxiliary Space: O(1)