# Count of root to leaf paths in a Binary Tree that form an AP

Given a Binary Tree, the task is to count all paths from root to leaf which forms an Arithmetic Progression.

Examples:

Input: Output:
Explanation:
The paths that form an AP in the given tree from root to leaf are:

• 1->3->5 (A.P. with common difference 2)

• 1->6->11 (A.P. with common difference 5)

Input: Output:
Explanation:
The path that form an AP in the given tree from root to leaf is 1->10->19 (A.P. with difference 9)

Approach: The problem can be solved using the Preorder Traversal. Follow the steps below to solve the problem:

• Perform Preorder Traversal on the given binary tree.
• Initialize an array arr[] to store the path.
• Initialize count = 0, to store the count of paths which forms an A.P.
• After reaching the leaf node, check if the current elements in the array(i.e. the node values from root to leaf path) forms an A.P..
• If so, increment the count
• After the complete traversal of the tree, print the count.

Below is the implementation of above approach:

## C++

 `// C++ implementation to count  ` `// the path which forms an A.P.  ` `#include   ` `using` `namespace` `std;  ` ` `  `int` `count = 0;  ` ` `  `// Node structure  ` `struct` `Node {  ` `    ``int` `val;  ` `    ``// left and right child of the node  ` `    ``Node *left, *right;  ` `    ``// intilizaiton consutructor  ` `    ``Node(``int` `x)  ` `    ``{  ` `        ``val = x;  ` `        ``left = NULL;  ` `        ``right = NULL;  ` `    ``}  ` `};  ` ` `  `// Function to check if path  ` `// forma A.P. or not  ` `bool` `check(vector<``int``> arr)  ` `{  ` ` `  `    ``if` `(arr.size() == 1)  ` `        ``return` `true``;  ` ` `  `    ``// if size of arr is greater than 2  ` `    ``int` `d = arr - arr;  ` ` `  `    ``for` `(``int` `i = 2; i < arr.size(); i++) {  ` `        ``if` `(arr[i] - arr[i - 1] != d)  ` `            ``return` `false``;  ` `    ``}  ` ` `  `    ``return` `true``;  ` `}  ` ` `  `// Function to find the maxmimum  ` `// setbits sum from root to leaf  ` `int` `countAP(Node* root, vector<``int``> arr)  ` `{  ` `    ``if` `(!root)  ` `        ``return` `0;  ` ` `  `    ``arr.push_back(root->val);  ` ` `  `    ``// If the node is a leaf node  ` `    ``if` `(root->left == NULL  ` `        ``&& root->right == NULL) {  ` `        ``if` `(check(arr))  ` `            ``return` `1;  ` `        ``return` `0;  ` `    ``}  ` ` `  `    ``// Traverse left subtree  ` `    ``int` `x = countAP(root->left, arr);  ` ` `  `    ``// Traverse the right subtree  ` `    ``int` `y = countAP(root->right, arr);  ` ` `  `    ``return` `x + y;  ` `}  ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` `    ``Node* root = ``new` `Node(1);  ` `    ``root->left = ``new` `Node(3);  ` `    ``root->right = ``new` `Node(6);  ` `    ``root->left->left = ``new` `Node(5);  ` `    ``root->left->right = ``new` `Node(7);  ` `    ``root->right->left = ``new` `Node(11);  ` `    ``root->right->right = ``new` `Node(23);  ` ` `  `    ``cout << countAP(root, {});  ` ` `  `    ``return` `0;  ` `}  `

## Python3

 `# Python3 implementation to count  ` `# the path which forms an A.P. ` ` `  `# Node structure ` `class` `Node: ` `    ``def` `__init__(``self``, x): ` `         `  `        ``self``.val ``=` `x ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` `         `  `# Function to check if path  ` `# form a A.P. or not  ` `def` `check(arr): ` `     `  `    ``if` `len``(arr) ``=``=` `1``: ` `        ``return` `True` `     `  `    ``# If size of arr is greater than 2  ` `    ``d ``=` `arr[``1``] ``-` `arr[``0``] ` `     `  `    ``for` `i ``in` `range``(``2``, ``len``(arr)): ` `        ``if` `arr[i] ``-` `arr[i ``-` `1``] !``=` `d: ` `            ``return` `False` `             `  `    ``return` `True` ` `  `# Function to find the maxmimum  ` `# setbits sum from root to leaf  ` `def` `countAP(root, arr): ` `     `  `    ``if` `not` `root: ` `        ``return` `0` `     `  `    ``arr.append(root.val) ` `     `  `    ``# If the node is a leaf node ` `    ``if` `(root.left ``=``=` `None` `and`  `        ``root.right ``=``=` `None``): ` `        ``if` `check(arr): ` `            ``return` `1` `        ``return` `0` `     `  `    ``# Traverse the left subtree ` `    ``x ``=` `countAP(root.left, arr) ` `     `  `    ``# Traverse the right subtree ` `    ``y ``=` `countAP(root.right, arr) ` `     `  `    ``return` `x ``+` `y ` ` `  `# Driver code ` `root ``=` `Node(``1``) ` `root.left ``=` `Node(``3``) ` `root.right ``=` `Node(``6``) ` `root.left.left ``=` `Node(``5``) ` `root.left.right ``=` `Node(``7``) ` `root.right.left ``=` `Node(``11``) ` `root.right.right ``=` `Node(``23``) ` ` `  `print``(countAP(root, [])) ` ` `  `# This code is contributed by stutipathak31jan `

Output:

`2`

Time Complexity: O(N)
Auxiliary Space: O(h), where h is the height of binary tree.

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