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Generate an array representing GCD of nodes of each vertical level of a Binary Tree
• Last Updated : 28 May, 2021

Given a Binary Tree, the task is to construct an array such that ith index of the array contains GCD of all the nodes present at the ith vertical level of the given binary tree.

Examples:

Input: Below is the given Tree:
5
/   \
4       7
/   \        \
8    10       6
\
8

Output: {8, 4, 5, 7, 6}

Explanation:
Vertical level I -> GCD(8) = 8.
Vertical level II -> GCD(4, 8) = 4.
Vertical level II -> GCD(5, 10) = 5.
Vertical level IV -> GCD(7) = 7.
Verticleal level V -> GCD(6) = 6

Input: Below is the given Tree:
4
/    \
2       3
/   \    /   \
3     2 4     5

Output: {3, 2, 2, 3, 5}

Approach: The given problem can be solved by performing the vertical order traversal of the given tree and stores each node’s value that occurs in the same vertical lines and then print the GCD of all the values stored for each level. Follow the below steps to solve this problem:

• Initialize a Map M to store the GCD of all the nodes for each vertical line while performing the traversal.
• Initialize a variable, say hd as 0 to keep track of the horizontal distance.
• Recursively traverse the given tree and perform the following steps:
• Store the current node’s values in the map M as the key as hd and value as node’s value.
• Decrement the variable hd by 1 and recursively call for the left subtree.
• Increment the variable hd by 1 and recursively call for the right subtree.
• Traverse the map and find the GCD of all the nodes stored in the map for each horizontal distance as the key and print the GCD value obtained.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Class for node of binary tree``struct` `TreeNode``{``    ``int` `val;``    ``TreeNode *left, *right;``    ` `    ``TreeNode(``int` `x)``    ``{``        ``val = x;``        ``left = right = NULL;``    ``}``};` `// Function to find GCD of two numbers``int` `GCD(``int` `a, ``int` `b)``{``    ``if` `(!b)``        ``return` `a;``        ` `    ``// Recursively find the GCD``    ``return` `GCD(b, a % b);``}` `// Stores the element for each``// vertical distance``unordered_map<``int``, ``int``> mp;` `// Function to traverse the tree``void` `Trav(TreeNode *root, ``int` `hd)``{``    ``if` `(!root)``        ``return``;` `    ``// Store the values in the map``    ``if` `(mp[hd] == 0)``        ``mp[hd] = root->val;``    ``else``        ``mp[hd] = GCD(mp[hd], root->val);` `    ``// Recursive Calls``    ``Trav(root->left, hd - 1);``    ``Trav(root->right, hd + 1);``}` `// Function to construct array from``// vertically positioned nodes``void` `constructArray(TreeNode *root)``{``    ``Trav(root, 0);``    ` `    ``// Get range of horizontal distances``    ``int` `lower = INT_MAX, upper = 0;``    ``for``(``auto` `it:mp)``    ``{``        ``lower = min(lower, it.first);``        ``upper = max(upper, it.first);``    ``}``    ` `    ``vector<``int``> ans;``    ` `    ``// Extract the array of values``    ``for``(``int` `i = lower; i < upper + 1; i++)``        ``ans.push_back(mp[i]);``    ` `    ``// Print the constructed array``    ``cout << ``"["``;``    ``for``(``int` `i = 0; i < ans.size() - 1; i++)``        ``cout << ans[i] << ``", "``;``        ` `    ``cout << ans[ans.size() - 1] << ``"]"``;``}` `// Driver code``int` `main()``{``    ``TreeNode *root = ``new` `TreeNode(5);``    ``root->left = ``new` `TreeNode(4);``    ``root->right = ``new` `TreeNode(7);``    ``root->left->left = ``new` `TreeNode(8);``    ``root->left->left->right = ``new` `TreeNode(8);``    ``root->left->right = ``new` `TreeNode(10);``    ``root->right->right = ``new` `TreeNode(6);``    ` `    ``// Function Call``    ``constructArray(root);``    ` `    ``return` `0;``}` `// This code is contributed by mohit kumar 29`

## Java

 `// Java program for the above approach``import` `java.lang.*;``import` `java.util.*;` `// Class containing the left and right``// child of current node and the``// key value``class` `Node``{``    ``int` `val;``    ``Node left, right;`` ` `    ``// Constructor of the class``    ``public` `Node(``int` `item)``    ``{``        ``val = item;``        ``left = right = ``null``;``    ``}``}` `class` `GFG{``    ` `Node root;` `// Stores the element for each``// vertical distance``static` `Map mp = ``new` `HashMap<>();`` ` `// Function to find GCD of two numbers``static` `int` `GCD(``int` `a, ``int` `b)``{``    ``if` `(b == ``0``)``        ``return` `a;``        ` `    ``// Recursively find the GCD``    ``return` `GCD(b, a % b);``}` `// Function to traverse the tree``static` `void` `Trav(Node root, ``int` `hd)``{``    ``if` `(root == ``null``)``        ``return``;` `    ``// Store the values in the map``    ``if` `(!mp.containsKey(hd))``        ``mp.put(hd, root.val);``    ``else``        ``mp.put(hd, GCD(mp.get(hd), root.val));` `    ``// Recursive Calls``    ``Trav(root.left, hd - ``1``);``    ``Trav(root.right, hd + ``1``);``}` `// Function to construct array from``// vertically positioned nodes``static` `void` `constructArray(Node root)``{``    ``Trav(root, ``0``);``    ` `    ``// Get range of horizontal distances``    ``int` `lower = Integer.MAX_VALUE, upper = ``0``;``    ``for``(Map.Entry it:mp.entrySet())``    ``{``        ``lower = Math.min(lower, it.getKey());``        ``upper = Math.max(upper, it.getKey());``    ``}``    ` `    ``ArrayList ans = ``new` `ArrayList<>();``    ` `    ``// Extract the array of values``    ``for``(``int` `i = lower; i < upper + ``1``; i++)``        ``ans.add(mp.getOrDefault(i,``0``));``    ` `    ``// Print the constructed array``    ``System.out.print(``"["``);``    ``for``(``int` `i = ``0``; i < ans.size() - ``1``; i++)``        ``System.out.print(ans.get(i) + ``", "``);``        ` `    ``System.out.print(ans.get(ans.size() - ``1``) + ``"]"``);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``GFG tree = ``new` `GFG();``    ` `    ``Node root = ``new` `Node(``5``);``    ``root.left = ``new` `Node(``4``);``    ``root.right = ``new` `Node(``7``);``    ``root.left.left = ``new` `Node(``8``);``    ``root.left.left.right = ``new` `Node(``8``);``    ``root.left.right = ``new` `Node(``10``);``    ``root.right.right = ``new` `Node(``6``);``    ` `    ``// Function Call``    ``constructArray(root);``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program for the above approach` `# Class for node of binary tree``class` `TreeNode:``    ``def` `__init__(``self``, val ``=``'', left ``=` `None``, right ``=` `None``):``        ``self``.val ``=` `val``        ``self``.left ``=` `left``        ``self``.right ``=` `right` `# Function to find GCD of two numbers``def` `GCD(a, b):``    ``if` `not` `b:``        ``return` `a` `    ``# Recursively find the GCD``    ``return` `GCD(b, a ``%` `b)` `# Function to construct array from``# vertically positioned nodes``def` `constructArray(root):``  ` `    ``# Stores the element for each``    ``# vertical distance``    ``mp ``=` `{}` `    ``# Function to traverse the tree``    ``def` `Trav(root, hd):``        ``if` `not` `root:``            ``return` `        ``# Store the values in the map``        ``if` `hd ``not` `in` `mp:``            ``mp[hd] ``=` `root.val``        ``else``:``            ``mp[hd] ``=` `GCD(mp[hd], root.val)` `        ``# Recursive Calls``        ``Trav(root.left, hd``-``1``)``        ``Trav(root.right, hd ``+` `1``)` `    ``Trav(root, ``0``)` `    ``# Get range of horizontal distances``    ``lower ``=` `min``(mp.keys())``    ``upper ``=` `max``(mp.keys())` `    ``ans ``=` `[]` `    ``# Extract the array of values``    ``for` `i ``in` `range``(lower, upper ``+` `1``):``        ``ans.append(mp[i])` `    ``# Print the constructed array``    ``print``(ans)`  `# Driver Code` `# Given Tree``root ``=` `TreeNode(``5``)``root.left ``=` `TreeNode(``4``)``root.right ``=` `TreeNode(``7``)``root.left.left ``=` `TreeNode(``8``)``root.left.left.right ``=` `TreeNode(``8``)``root.left.right ``=` `TreeNode(``10``)``root.right.right ``=` `TreeNode(``6``)` `# Function Call``constructArray(root)`
Output:
`[8, 4, 5, 7, 6]`

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

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