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SumRootToLeaf.java
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SumRootToLeaf.java
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/*
Sum up root to the leaf number
You are provided with the roots of the binary tree
that contains digits from 0 to 9 only.
Each root to leaf path in the tree represents a number.
For example :-
1
/ \
2 3
has 12 and 13, as numbers forming from the tree.
Your task is to return the total sum of all root to leaf numbers.
*/
import java.io.*;
import java.until.*;
//defining a class for storing nodes of tree
class BTNode
{
int value;
BTNode left;
BTNode right;
BTNode()
{}
BTNode(int value)
{
this.value = value;
}
BTNode(int value, BTNode left, BTNode right)
{
this.value = value;
this.left = left;
this.right = right;
}
}
public class SumRootToLeaf
{
//initializing a variable named result to store the final result
int result;
//function that sums up the root-to-leaf numbers
public int sumNumb(BTNode root)
{
//if the root is empty the result returned is zero
if(root == null)
{
return 0;
}
result = 0;
findSum(root, 0);
return result;
}
//function to find sum
public void findSum(BTNode root, int value)
{
int currentval = value * 10 + root.value;
//if both the sides of the tree are containing null then
//current number formed is added to the result and is simply
//returned
if(root.left == null && root.right == null)
{
result = result + currentval;
return;
}
if(root.left != null)
{
findSum(root.left, currentval);
}
if(root.right != null)
{
findSum(root.right, currentval);
}
}
//driver code
public static void main()
{
// Taking input
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the values of nodes for tree : ");
String str = br.readLine();
String ip[] = str.split(" ");
// Here we start creating the root of the tree
Node root = new Node(Integer.parseInt(ip[0]));
// Pushing the roots to the queue
Queue<Node> Treequeue = new LinkedList<>();
Treequeue.add(root);
//for output
System.out.println("Sum of the root-to-leaf path is : ");
System.out.print(SumRootToLeaf(root));
System.out.println(" ");
}
}
/*
EXAMPLE:-
INPUT--
Enter the values of nodes for tree : 4 9 0 5 1
OUTPUT--
Sum of the root-to-leaf path is : 1026
4
/ \
9 0
/ \
5 1
has numbers formed by root-to-leaf iteration is
495, 491, 40.
So, the sum comes out to be 495 + 491 + 40 = 1026.
TIME COMPLEXITY -- > O(N)
SPACE COMPLEXITY --> O(N) ; where n is no. of roots
*/