Give an algorithm for printing the level order data in reverse order

For example, the output for the below tree should be: 4 5 6 7 2 3 1

public class PrintDataInReverseOrder {
    public static void main(String args[]) {
             Node rootNode = new Node("1");
             rootNode.left = new Node("2");
             rootNode.left.left = new Node("4");
             rootNode.left.right = new Node("5");
             rootNode.right = new Node("3");
             rootNode.right.right = new Node("7");
             rootNode.right.left = new Node("6");
           
             printInreverse(rootNode);
    }
   
    static void printInreverse(Node node) {
       
        Queue<Node> queue = new LinkedBlockingQueue<>();
        queue.add(node);
        Stack<String> stack = new Stack<>();
        while(!queue.isEmpty()) {
            Node node1 = queue.remove();
            stack.add(node1.value);
            if(null!=node1.right) {
                queue.add(node1.right);
            }
            if(null!=node1.left) {
                queue.add(node1.left);
            }
        }
       
        while(!stack.isEmpty()) {
            System.out.print(stack.pop());
        }
    }
}

class BinayTree{
    static class Node{
        String value;
        Node left;
        Node right;
       
        Node(String val){
            value = val;
            left = right = null;
        }
       
    }
   
}

output: 4567231

Calculate Size of binary tree Using Recusrion

public class CalculateSizeUsingRecusrion {
    public static void main(String args[]) {
             Node rootNode = new Node("4");
             rootNode.left = new Node("2");
             rootNode.left.left = new Node("1");
             rootNode.left.right = new Node("3");
             rootNode.right = new Node("6");
             rootNode.right.right = new Node("7");
             rootNode.right.left = new Node("5");
           
             System.out.println(calcSize(rootNode));
    }
   
    static int calcSize(Node node) {
        if(node==null) {
            return 0;
        }
        return calcSize(node.left)+1+calcSize(node.right);
    }
}

class BinayTree{
    static class Node{
        String value;
        Node left;
        Node right;
       
        Node(String val){
            value = val;
            left = right = null;
        }
       
    }
   
}

output: 7

Calculate Size of binary tree Using Without Recusrion

public class CalculateSizeUsingWithoutRecusrion {
    public static void main(String args[]) {
             Node rootNode = new Node("4");
             rootNode.left = new Node("2");
             rootNode.left.left = new Node("1");
             rootNode.left.left.left = new Node("9");
             rootNode.left.right = new Node("3");
             rootNode.right = new Node("6");
             rootNode.right.right = new Node("7");
             rootNode.right.left = new Node("5");
             rootNode.right.right.right = new Node("8");
           
             System.out.println(calcSize(rootNode));
    }
   
    static int calcSize(Node node) {
        Queue<Node> queue = new LinkedBlockingQueue<>();
        queue.add(node);
        int count = 0;
        while(!queue.isEmpty()) {
            Node node1 = queue.remove();
            count= count+1;
            if(null!=node1.left) {
                queue.add(node1.left);
            }
            if(null!=node1.right) {
                queue.add(node1.right);
            }
        }
        return count;
    }
}



class BinayTree{
    static class Node{
        String value;
        Node left;
        Node right;
       
        Node(String val){
            value = val;
            left = right = null;
        }
       
    }
   
}


output:9

How to revert merged commit in git

Assume i have a branch master and i have done multiple commit to that branch till  1e3a87c,
now create a new branch from master is dev , push some  commit to dev branch till b038437,

now merge dev branch changes into master as below:

# checkout to master branch
# git pull origin dev
# git push origin master

and if you want to merge with tag as below:
#create a tag on dev branch using git tag tag_name  
# checkout to master branch
#git merge tag_name
#git push origin master

Now dev branch merge into master branch

As if we want to revert commit before merge then need to use below:

we need to find out the commit id before merge like 1e3a87c in my case


#git reset 1e3a87c
#git reset --hard
#git push -f origin master


Ref: https://stackoverflow.com/questions/7099833/how-to-revert-a-merge-commit-thats-already-pushed-to-remote-branch/7100005#7100005

PrintExcelColumnNameFromColumnNumber

public class PrintExcelColumnName {
    public static void main(String args[]) {
        printColumnName(20000);
    }
   
    static void printColumnName(int num){
        StringBuilder sb = new StringBuilder();
        while(num>26) {
            int diff = num%26;
            num = num/26;
            sb.append((char)(diff-1+'A'));
        }
        sb.append((char)(num-1+'A'));
       
        System.out.println(sb.reverse().toString());
    }
   
   
}

output: ACOF

FindUniqueDigitsNumber from list

public class FindUniqueDigitsNumber {
    public static void main(String args[]) {
        int l = 10;
        int r = 30;
        findUniqueDigitNumber(l, r);
       
    }
   
    static void findUniqueDigitNumber(int l, int r){
        for(int i=l;i<=r;i++) {
            boolean[] boo = new boolean[100];
            int num = i;
            while(num!=0) {
                if(boo[num%10]) {
                    break;
                }
                boo[num%10] = true;
                num = num/10;
            }
            if(num==0)
            {
                System.out.println(i);
            }
           
        }
    }
   
}

output(11 and 22 missing):
10
12
13
14
15
16
17
18
19
20
21
23
24
25
26
27
28
29
30

FindMissing number in AP log(n)

public class FindMissing {

    public static void main(String[] args) {
        int[] arr = {2,4,6,8,12,14,16,18,20,22,24,26};
        int val = findMissing(arr, 0, arr.length-1, 2);
        System.out.println(val);
    }
   
   
    static int findMissing(int[] arr, int start, int end,int diff) {
        if (end >= start) {
            int mid = (start + end) / 2;
            if (mid<end && (arr[mid+1] - arr[mid]) != diff) {
                return arr[mid] + diff;
            } else if (mid>0 && (arr[mid] - arr[mid-1]) != diff) {
                return arr[mid-1] + diff;
            }
//if difference in left and right element of mid are equal to diff then we need to decide to move either //left or right as we will check if our series is correct till mid then we will move to right otherwise left
//for checking till mid series  is correct we will use formula : arr[mid]==arr[0]+mid*diff


else if (arr[mid] == arr[0] + mid * diff) {
                return findMissing(arr, mid + 1, end, diff);
            } else {
                return findMissing(arr, start, mid - 1, diff);
            }

        }
        return 0;
    }

}