Generate an array representing GCD of nodes of each vertical level of a Binary Tree
Last Updated :
01 Jul, 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 <bits/stdc++.h>
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 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<Integer, Integer> 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<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 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 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)
// 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 value
public 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 distance
static Dictionary<int,
int> mp = new Dictionary<int,
int>();
// 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.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 nodes
static 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 code
static 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
<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>
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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