# Maximum Consecutive Increasing Path Length in Binary Tree

Given a Binary Tree find the length of the longest path which comprises of nodes with consecutive values in increasing order. Every node is considered as a path of length 1.

Examples:

10 / \ / \ 11 9 / \ /\ / \ / \ 13 12 13 8 Maximum Consecutive Path Length is 3 (10, 11, 12)Note: 10, 9 ,8 is NOT considered since the nodes should be in increasing order. 5 / \ / \ 8 11 / \ / \ 9 10 / / / / 6 15 Maximum Consecutive Path Length is 2 (8, 9).

Every node in the Binary Tree can either become part of the path which is starting from one of its parent node or a new path can start from this node itself. The key is to recursively find the path length for the left and right sub tree and then return the maximum. Some cases need to be considered while traversing the tree which are discussed below.*prev *: stores the value of the parent node. Initialize prev with one less than value of root node so that the path starting at root can be of length at least 1. *len *: Stores the path length which ends at the parent of currently visited node.**Case 1**: Value of Current Node is prev +1

In this case increase the path length by 1, and then recursively find the path length for the left and the right sub tree then return the maximum between two lengths.**Case 2**: Value of Current Node is NOT prev+1

A new path can start from this node, so recursively find the path length for the left and the right sub tree. The path which ends at the parent node of current node might be greater than the path which starts from this node.So take the maximum of the path which starts from this node and which ends at previous node.

Below is the implementation of above idea.

## C++

`// C++ Program to find Maximum Consecutive` `// Path Length in a Binary Tree` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// To represent a node of a Binary Tree` `struct` `Node` `{` ` ` `Node *left, *right;` ` ` `int` `val;` `};` `// Create a new Node and return its address` `Node *newNode(` `int` `val)` `{` ` ` `Node *temp = ` `new` `Node();` ` ` `temp->val = val;` ` ` `temp->left = temp->right = NULL;` ` ` `return` `temp;` `}` `// Returns the maximum consecutive Path Length` `int` `maxPathLenUtil(Node *root, ` `int` `prev_val, ` `int` `prev_len)` `{` ` ` `if` `(!root)` ` ` `return` `prev_len;` ` ` `// Get the value of Current Node` ` ` `// The value of the current node will be` ` ` `// prev Node for its left and right children` ` ` `int` `cur_val = root->val;` ` ` `// If current node has to be a part of the` ` ` `// consecutive path then it should be 1 greater` ` ` `// than the value of the previous node` ` ` `if` `(cur_val == prev_val+1)` ` ` `{` ` ` `// a) Find the length of the Left Path` ` ` `// b) Find the length of the Right Path` ` ` `// Return the maximum of Left path and Right path` ` ` `return` `max(maxPathLenUtil(root->left, cur_val, prev_len+1),` ` ` `maxPathLenUtil(root->right, cur_val, prev_len+1));` ` ` `}` ` ` `// Find length of the maximum path under subtree rooted with this` ` ` `// node (The path may or may not include this node)` ` ` `int` `newPathLen = max(maxPathLenUtil(root->left, cur_val, 1),` ` ` `maxPathLenUtil(root->right, cur_val, 1));` ` ` `// Take the maximum previous path and path under subtree rooted` ` ` `// with this node.` ` ` `return` `max(prev_len, newPathLen);` `}` `// A wrapper over maxPathLenUtil().` `int` `maxConsecutivePathLength(Node *root)` `{` ` ` `// Return 0 if root is NULL` ` ` `if` `(root == NULL)` ` ` `return` `0;` ` ` `// Else compute Maximum Consecutive Increasing Path` ` ` `// Length using maxPathLenUtil.` ` ` `return` `maxPathLenUtil(root, root->val-1, 0);` `}` `//Driver program to test above function` `int` `main()` `{` ` ` `Node *root = newNode(10);` ` ` `root->left = newNode(11);` ` ` `root->right = newNode(9);` ` ` `root->left->left = newNode(13);` ` ` `root->left->right = newNode(12);` ` ` `root->right->left = newNode(13);` ` ` `root->right->right = newNode(8);` ` ` `cout << ` `"Maximum Consecutive Increasing Path Length is "` ` ` `<< maxConsecutivePathLength(root);` ` ` `return` `0;` `}` |

## Java

`// Java Program to find Maximum Consecutive` `// Path Length in a Binary Tree` `import` `java.util.*;` `class` `GfG {` `// To represent a node of a Binary Tree` `static` `class` `Node` `{` ` ` `Node left, right;` ` ` `int` `val;` `}` `// Create a new Node and return its address` `static` `Node newNode(` `int` `val)` `{` ` ` `Node temp = ` `new` `Node();` ` ` `temp.val = val;` ` ` `temp.left = ` `null` `;` ` ` `temp.right = ` `null` `;` ` ` `return` `temp;` `}` `// Returns the maximum consecutive Path Length` `static` `int` `maxPathLenUtil(Node root, ` `int` `prev_val, ` `int` `prev_len)` `{` ` ` `if` `(root == ` `null` `)` ` ` `return` `prev_len;` ` ` `// Get the value of Current Node` ` ` `// The value of the current node will be` ` ` `// prev Node for its left and right children` ` ` `int` `cur_val = root.val;` ` ` `// If current node has to be a part of the` ` ` `// consecutive path then it should be 1 greater` ` ` `// than the value of the previous node` ` ` `if` `(cur_val == prev_val+` `1` `)` ` ` `{` ` ` `// a) Find the length of the Left Path` ` ` `// b) Find the length of the Right Path` ` ` `// Return the maximum of Left path and Right path` ` ` `return` `Math.max(maxPathLenUtil(root.left, cur_val, prev_len+` `1` `),` ` ` `maxPathLenUtil(root.right, cur_val, prev_len+` `1` `));` ` ` `}` ` ` `// Find length of the maximum path under subtree rooted with this` ` ` `// node (The path may or may not include this node)` ` ` `int` `newPathLen = Math.max(maxPathLenUtil(root.left, cur_val, ` `1` `),` ` ` `maxPathLenUtil(root.right, cur_val, ` `1` `));` ` ` `// Take the maximum previous path and path under subtree rooted` ` ` `// with this node.` ` ` `return` `Math.max(prev_len, newPathLen);` `}` `// A wrapper over maxPathLenUtil().` `static` `int` `maxConsecutivePathLength(Node root)` `{` ` ` `// Return 0 if root is NULL` ` ` `if` `(root == ` `null` `)` ` ` `return` `0` `;` ` ` `// Else compute Maximum Consecutive Increasing Path` ` ` `// Length using maxPathLenUtil.` ` ` `return` `maxPathLenUtil(root, root.val-` `1` `, ` `0` `);` `}` `//Driver program to test above function` `public` `static` `void` `main(String[] args)` `{` ` ` `Node root = newNode(` `10` `);` ` ` `root.left = newNode(` `11` `);` ` ` `root.right = newNode(` `9` `);` ` ` `root.left.left = newNode(` `13` `);` ` ` `root.left.right = newNode(` `12` `);` ` ` `root.right.left = newNode(` `13` `);` ` ` `root.right.right = newNode(` `8` `);` ` ` `System.out.println(` `"Maximum Consecutive Increasing Path Length is "` `+maxConsecutivePathLength(root));` `}` `}` |

## Python

`# Python program to find Maximum consecutive` `# path length in binary tree` `# A binary tree node` `class` `Node:` ` ` ` ` `# Constructor to create a new node` ` ` `def` `__init__(` `self` `, val):` ` ` `self` `.val ` `=` `val` ` ` `self` `.left ` `=` `None` ` ` `self` `.right ` `=` `None` `# Returns the maximum consecutive path length` `def` `maxPathLenUtil(root, prev_val, prev_len):` ` ` `if` `root ` `is` `None` `:` ` ` `return` `prev_len` ` ` `# Get the vlue of current node` ` ` `# The value of the current node will be` ` ` `# prev node for its left and right children` ` ` `curr_val ` `=` `root.val` ` ` ` ` `# If current node has to be a part of the` ` ` `# consecutive path then it should be 1 greater` ` ` `# thn the value of the previous node` ` ` `if` `curr_val ` `=` `=` `prev_val ` `+` `1` `:` ` ` ` ` `# a) Find the length of the left path` ` ` `# b) Find the length of the right path` ` ` `# Return the maximum of left path and right path` ` ` `return` `max` `(maxPathLenUtil(root.left, curr_val, prev_len` `+` `1` `),` ` ` `maxPathLenUtil(root.right, curr_val, prev_len` `+` `1` `))` ` ` `# Find the length of the maximum path under subtree` ` ` `# rooted with this node` ` ` `newPathLen ` `=` `max` `(maxPathLenUtil(root.left, curr_val, ` `1` `),` ` ` `maxPathLenUtil(root.right, curr_val, ` `1` `))` ` ` `# Take the maximum previous path and path under subtree` ` ` `# rooted with this node` ` ` `return` `max` `(prev_len , newPathLen)` `# A Wrapper over maxPathLenUtil()` `def` `maxConsecutivePathLength(root):` ` ` ` ` `# Return 0 if root is None` ` ` `if` `root ` `is` `None` `:` ` ` `return` `0` ` ` ` ` `# Else compute maximum consecutive increasing path` ` ` `# length using maxPathLenUtil` ` ` `return` `maxPathLenUtil(root, root.val ` `-` `1` `, ` `0` `)` `# Driver program to test above function` `root ` `=` `Node(` `10` `)` `root.left ` `=` `Node(` `11` `)` `root.right ` `=` `Node(` `9` `)` `root.left.left ` `=` `Node(` `13` `)` `root.left.right ` `=` `Node(` `12` `)` `root.right.left ` `=` `Node(` `13` `)` `root.right.right ` `=` `Node(` `8` `)` `print` `"Maximum Consecutive Increasing Path Length is"` `,` `print` `maxConsecutivePathLength(root)` `# This code is contributed by Nikhil Kumar Singh(nickzuck_007)` |

## C#

`// C# Program to find Maximum Consecutive` `// Path Length in a Binary Tree` `using` `System;` `class` `GfG` `{` ` ` `// To represent a node of a Binary Tree` ` ` `class` `Node` ` ` `{` ` ` `public` `Node left, right;` ` ` `public` `int` `val;` ` ` `}` ` ` `// Create a new Node and return its address` ` ` `static` `Node newNode(` `int` `val)` ` ` `{` ` ` `Node temp = ` `new` `Node();` ` ` `temp.val = val;` ` ` `temp.left = ` `null` `;` ` ` `temp.right = ` `null` `;` ` ` `return` `temp;` ` ` `}` ` ` `// Returns the maximum consecutive Path Length` ` ` `static` `int` `maxPathLenUtil(Node root,` ` ` `int` `prev_val, ` `int` `prev_len)` ` ` `{` ` ` `if` `(root == ` `null` `)` ` ` `return` `prev_len;` ` ` `// Get the value of Current Node` ` ` `// The value of the current node will be` ` ` `// prev Node for its left and right children` ` ` `int` `cur_val = root.val;` ` ` `// If current node has to be a part of the` ` ` `// consecutive path then it should be 1 greater` ` ` `// than the value of the previous node` ` ` `if` `(cur_val == prev_val+1)` ` ` `{` ` ` `// a) Find the length of the Left Path` ` ` `// b) Find the length of the Right Path` ` ` `// Return the maximum of Left path and Right path` ` ` `return` `Math.Max(maxPathLenUtil(root.left, cur_val, prev_len+1),` ` ` `maxPathLenUtil(root.right, cur_val, prev_len+1));` ` ` `}` ` ` `// Find length of the maximum path under subtree rooted with this` ` ` `// node (The path may or may not include this node)` ` ` `int` `newPathLen = Math.Max(maxPathLenUtil(root.left, cur_val, 1),` ` ` `maxPathLenUtil(root.right, cur_val, 1));` ` ` `// Take the maximum previous path and path under subtree rooted` ` ` `// with this node.` ` ` `return` `Math.Max(prev_len, newPathLen);` ` ` `}` ` ` `// A wrapper over maxPathLenUtil().` ` ` `static` `int` `maxConsecutivePathLength(Node root)` ` ` `{` ` ` `// Return 0 if root is NULL` ` ` `if` `(root == ` `null` `)` ` ` `return` `0;` ` ` `// Else compute Maximum Consecutive Increasing Path` ` ` `// Length using maxPathLenUtil.` ` ` `return` `maxPathLenUtil(root, root.val - 1, 0);` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main(String[] args)` ` ` `{` ` ` `Node root = newNode(10);` ` ` `root.left = newNode(11);` ` ` `root.right = newNode(9);` ` ` `root.left.left = newNode(13);` ` ` `root.left.right = newNode(12);` ` ` `root.right.left = newNode(13);` ` ` `root.right.right = newNode(8);` ` ` `Console.WriteLine(` `"Maximum Consecutive"` `+` ` ` `" Increasing Path Length is "` `+` ` ` `maxConsecutivePathLength(root));` ` ` `}` `}` `// This code has been contributed by 29AjayKumar` |

## Javascript

`<script>` `// Javascript Program to find Maximum Consecutive` `// Path Length in a Binary Tree` `// To represent a node of a Binary Tree` `class Node` `{` ` ` `constructor(val)` ` ` `{` ` ` `this` `.val = val;` ` ` `this` `.left = ` `this` `.right = ` `null` `;` ` ` `}` `}` `// Returns the maximum consecutive Path Length` `function` `maxPathLenUtil(root,prev_val,prev_len)` `{` ` ` `if` `(root == ` `null` `)` ` ` `return` `prev_len;` ` ` ` ` `// Get the value of Current Node` ` ` `// The value of the current node will be` ` ` `// prev Node for its left and right children` ` ` `let cur_val = root.val;` ` ` ` ` `// If current node has to be a part of the` ` ` `// consecutive path then it should be 1 greater` ` ` `// than the value of the previous node` ` ` `if` `(cur_val == prev_val+1)` ` ` `{` ` ` ` ` `// a) Find the length of the Left Path` ` ` `// b) Find the length of the Right Path` ` ` `// Return the maximum of Left path and Right path` ` ` `return` `Math.max(maxPathLenUtil(root.left, cur_val, prev_len+1),` ` ` `maxPathLenUtil(root.right, cur_val, prev_len+1));` ` ` `}` ` ` ` ` `// Find length of the maximum path under subtree rooted with this` ` ` `// node (The path may or may not include this node)` ` ` `let newPathLen = Math.max(maxPathLenUtil(root.left, cur_val, 1),` ` ` `maxPathLenUtil(root.right, cur_val, 1));` ` ` ` ` `// Take the maximum previous path and path under subtree rooted` ` ` `// with this node.` ` ` `return` `Math.max(prev_len, newPathLen);` `}` `// A wrapper over maxPathLenUtil().` `function` `maxConsecutivePathLength(root)` `{` ` ` `// Return 0 if root is NULL` ` ` `if` `(root == ` `null` `)` ` ` `return` `0;` ` ` ` ` `// Else compute Maximum Consecutive Increasing Path` ` ` `// Length using maxPathLenUtil.` ` ` `return` `maxPathLenUtil(root, root.val-1, 0);` `}` `// Driver program to test above function` `let root = ` `new` `Node(10);` `root.left = ` `new` `Node(11);` `root.right = ` `new` `Node(9);` `root.left.left = ` `new` `Node(13);` `root.left.right = ` `new` `Node(12);` `root.right.left = ` `new` `Node(13);` `root.right.right = ` `new` `Node(8);` `document.write(` `"Maximum Consecutive Increasing Path Length is "` `+` `maxConsecutivePathLength(root)+` `"<br>"` `);` `// This code is contributed by rag2127` `</script>` |

**Output:**

Maximum Consecutive Increasing Path Length is 3

This article is contributed by **Chirag Agarwal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.