Diagonal Sum of a Binary Tree - GeeksforGeeks

Diagonal Sum of a Binary Tree - GeeksforGeeks
Diagonal Sum of a Binary Tree Consider lines of slope -1 passing between nodes (dotted lines in below diagram). Diagonal sum in a binary tree is sum of all node's data lying between these lines. Given a Binary Tree, print all diagonal sums.

For the following input tree, output should be 9, 19, 42.
9 is sum of 1, 3 and 5.
19 is sum of 2, 6, 4 and 7.
42 is sum of 9, 10, 11 and 12.

The idea is to keep track of vertical distance from top diagonal passing through root. We increment the vertical distance we go down to next diagonal.
1. Add root with vertical distance as 0 to the queue.
2. Process the sum of all right child and right of right child and so on.
3. Add left child current node into the queue for later processing. The vertical distance of left child is vertical distance of current node plus 1.
4. Keep doing 2nd, 3rd and 4th step till the queue is empty.

class TreeNode

{

    int data; //node data

    int vd; //vertical distance diagonally

    TreeNode left, right; //left and right child's reference

 

    // Tree node constructor

    public TreeNode(int data)

    {

        this.data = data;

        vd = Integer.MAX_VALUE;

        left = right = null;

    }

}

 

// Tree class

class Tree

{

    TreeNode root;//Tree root

 

    // Tree constructor

    public Tree(TreeNode root)  {  this.root = root;  }

 

    // Diagonal sum method

    public void diagonalSum()

    {

        // Queue which stores tree nodes

        Queue<TreeNode> queue = new LinkedList<TreeNode>();

 

        // Map to store sum of node's data lying diagonally

        Map<Integer, Integer> map = new TreeMap<>();

 

        // Assign the root's vertical distance as 0.

        root.vd = 0;

 

        // Add root node to the queue

        queue.add(root);

 

        // Loop while the queue is not empty

        while (!queue.isEmpty())

        {

            // Remove the front tree node from queue.

            TreeNode curr = queue.remove();

 

            // Get the vertical distance of the dequeued node.

            int vd = curr.vd;

 

            // Sum over this node's right-child, right-of-right-child

            // and so on

            while (curr != null)

            {

                int prevSum = (map.get(vd) == null)? 0: map.get(vd);

                map.put(vd, prevSum + curr.data);

 

                // If for any node the left child is not null add

                // it to the queue for future processing.

                if (curr.left != null)

                {

                    curr.left.vd = vd+1;

                    queue.add(curr.left);

                }

 

                // Move to the current node's right child.

                curr = curr.right;

            }

        }

 

        // Make an entry set from map.

        Set<Entry<Integer, Integer>> set = map.entrySet();

 

        // Make an iterator

        Iterator<Entry<Integer, Integer>> iterator = set.iterator();

 

        // Traverse the map elements using the iterator.

        while (iterator.hasNext())

        {

            Map.Entry<Integer, Integer> me = iterator.next();

            System.out.print(me.getValue()+" ");

        }

    }

}

http://www.ideserve.co.in/learn/diagonal-sum-of-a-binary-tree

1. Base case - if root is null return 0;
2. Otherwise, at the current node 'n' located at diagonal no. 'd'
    a. Compute the diagonal sum for the left sub-tree of 'n'. The left sub-tree diagonal would start with 'd+1'
    b. Add the current node n's value to sum of diagonal 'd' and keep traversing right.

This problem was for diagonals from top to bottom and slope -1. Try the same problem for slope +1.

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