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
\
8Output: {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) = 6Input: Below is the given Tree:
4
/ \
2 3
/ \ / \
3 2 4 5Output: {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 <bits/stdc++.h>using namespace std;// Class for node of binary treestruct TreeNode{ int val; TreeNode *left, *right; TreeNode(int x) { val = x; left = right = NULL; }};// Function to find GCD of two numbersint GCD(int a, int b){ if (!b) return a; // Recursively find the GCD return GCD(b, a % b);}// Stores the element for each// vertical distanceunordered_map<int, int> mp;// Function to traverse the treevoid 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 nodesvoid 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 codeint 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 approachimport java.lang.*;import java.util.*;// Class containing the left and right// child of current node and the// key valueclass 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 distancestatic Map<Integer, Integer> mp = new HashMap<>(); // Function to find GCD of two numbersstatic int GCD(int a, int b){ if (b == 0) return a; // Recursively find the GCD return GCD(b, a % b);}// Function to traverse the treestatic 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 nodesstatic void constructArray(Node root){ Trav(root, 0); // Get range of horizontal distances int lower = Integer.MAX_VALUE, upper = 0; for(Map.Entry<Integer, Integer> it:mp.entrySet()) { lower = Math.min(lower, it.getKey()); upper = Math.max(upper, it.getKey()); } ArrayList<Integer> 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 codepublic 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 treeclass TreeNode: def __init__(self, val ='', left = None, right = None): self.val = val self.left = left self.right = right# Function to find GCD of two numbersdef GCD(a, b): if not b: return a # Recursively find the GCD return GCD(b, a % b)# Function to construct array from# vertically positioned nodesdef 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 Treeroot = 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 CallconstructArray(root) |
C#
// C# program for the above approach using System;using System.Collections.Generic;// Class containing the left and right// child of current node and the// key valuepublic class Node{ public int val; public Node left, right; // Constructor of the class public Node(int item) { val = item; left = right = null; }}class GFG{ // Stores the element for each// vertical distancestatic Dictionary<int, int> mp = new Dictionary<int, int>(); // Function to find GCD of two numbersstatic int GCD(int a, int b){ if (b == 0) return a; // Recursively find the GCD return GCD(b, a % b);} // Function to traverse the treestatic void Trav(Node root, int hd){ if (root == null) return; // Store the values in the map if (!mp.ContainsKey(hd)) mp.Add(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 nodesstatic void constructArray(Node root){ Trav(root, 0); // Get range of horizontal distances int lower = Int32.MaxValue, upper = 0; foreach(KeyValuePair<int, int> it in mp) { lower = Math.Min(lower, it.Key); upper = Math.Max(upper, it.Key); } List<int> ans = new List<int>(); // Extract the array of values for(int i = lower; i < upper + 1; i++) if(mp.ContainsKey(i)) ans.Add(mp[i]); else ans.Add(0); // Print the constructed array Console.Write("["); for(int i = 0; i < ans.Count - 1; i++) Console.Write(ans[i] + ", "); Console.Write(ans[ans.Count - 1] + "]");} // Driver codestatic public void Main(){ 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 rishavmahato348 |
Javascript
<script> // Javascript program for the above approach // Class containing the left and right // child of current node and the // key value class Node { constructor(item) { this.left = null; this.right = null; this.val = item; } } // Stores the element for each // vertical distance let mp = new Map(); // Function to find GCD of two numbers function GCD(a, b) { if (b == 0) return a; // Recursively find the GCD return GCD(b, a % b); } // Function to traverse the tree function Trav(root, hd) { if (root == null) return; // Store the values in the map if (!mp.has(hd)) mp.set(hd, root.val); else mp.set(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 function constructArray(root) { Trav(root, 0); // Get range of horizontal distances let lower = Number.MAX_VALUE, upper = 0; mp.forEach((values,keys)=>{ lower = Math.min(lower, keys); upper = Math.max(upper, keys); }) let ans = []; // Extract the array of values for(let i = lower; i < upper + 1; i++) { if(mp.has(i)) { ans.push(mp.get(i)); } else{ ans.push(0); } } // Print the constructed array document.write("["); for(let i = 0; i < ans.length - 1; i++) document.write(ans[i] + ", "); document.write(ans[ans.length - 1] + "]"); } let 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 suresh07.</script> |
Output:
[8, 4, 5, 7, 6]
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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