# Diagonal Sum of a Binary Tree

Consider lines of slope -1 passing between nodes (dotted lines in below diagram). The diagonal sum in a binary tree is the sum of all node’s data lying between these lines. Given a Binary Tree, print all diagonal sums.
For the following input tree, the 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. ### Method 1:

Algorithm:
The idea is to keep track of vertical distance from top diagonal passing through the 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 the right child and so on.
3. Add left child current node into the queue for later processing. The vertical distance of the left child is the vertical distance of current node plus 1.
4. Keep doing 2nd, 3rd and 4th step till the queue is empty.

Following is the implementation of the above idea.

## C++

 `// C++ Program to find diagonal` `// sum in a Binary Tree` `#include ` `using` `namespace` `std;`   `struct` `Node` `{` `    ``int` `data;` `    ``struct` `Node* left;` `    ``struct` `Node* right;` `};`   `struct` `Node* newNode(``int` `data)` `{` `    ``struct` `Node* Node =` `            ``(``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));` `    `  `    ``Node->data = data;` `    ``Node->left = Node->right = NULL;`   `    ``return` `Node;` `}`   `// root - root of the binary tree` `// vd - vertical distance diagonally` `// diagonalSum - map to store Diagonal ` `// Sum(Passed by Reference)` `void` `diagonalSumUtil(``struct` `Node* root,` `                ``int` `vd, map<``int``, ``int``> &diagonalSum)` `{` `    ``if``(!root)` `        ``return``;` `        `  `    ``diagonalSum[vd] += root->data;`   `    ``// increase the vertical distance if left child` `    ``diagonalSumUtil(root->left, vd + 1, diagonalSum);`   `    ``// vertical distance remains same for right child` `    ``diagonalSumUtil(root->right, vd, diagonalSum);` `}`   `// Function to calculate diagonal ` `// sum of given binary tree` `void` `diagonalSum(``struct` `Node* root)` `{`   `    ``// create a map to store Diagonal Sum` `    ``map<``int``, ``int``> diagonalSum; ` `    `  `    ``diagonalSumUtil(root, 0, diagonalSum);`   `    ``map<``int``, ``int``>::iterator it;` `        ``cout << ``"Diagonal sum in a binary tree is - "``;` `    `  `    ``for``(it = diagonalSum.begin();` `                ``it != diagonalSum.end(); ++it)` `    ``{` `        ``cout << it->second << ``" "``;` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``struct` `Node* root = newNode(1);` `    ``root->left = newNode(2);` `    ``root->right = newNode(3);` `    ``root->left->left = newNode(9);` `    ``root->left->right = newNode(6);` `    ``root->right->left = newNode(4);` `    ``root->right->right = newNode(5);` `    ``root->right->left->right = newNode(7);` `    ``root->right->left->left = newNode(12);` `    ``root->left->right->left = newNode(11);` `    ``root->left->left->right = newNode(10);`   `    ``diagonalSum(root);`   `    ``return` `0;` `}`   `// This code is contributed by Aditya Goel`

## Java

 `// Java Program to find diagonal sum in a Binary Tree` `import` `java.util.*;` `import` `java.util.Map.Entry;`   `//Tree node` `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 queue = ``new` `LinkedList();`   `        ``// Map to store sum of node's data lying diagonally` `        ``Map 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> set = map.entrySet();`   `        ``// Make an iterator` `        ``Iterator> iterator = set.iterator();`   `        ``// Traverse the map elements using the iterator.` `         ``System.out.print(``"Diagonal sum in a binary tree is - "``);` `        ``while` `(iterator.hasNext())` `        ``{` `            ``Map.Entry me = iterator.next();`   `            ``System.out.print(me.getValue()+``" "``);` `        ``}` `    ``}` `}`   `//Driver class` `public` `class` `DiagonalSum` `{` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``TreeNode root = ``new` `TreeNode(``1``);` `        ``root.left = ``new` `TreeNode(``2``);` `        ``root.right = ``new` `TreeNode(``3``);` `        ``root.left.left = ``new` `TreeNode(``9``);` `        ``root.left.right = ``new` `TreeNode(``6``);` `        ``root.right.left = ``new` `TreeNode(``4``);` `        ``root.right.right = ``new` `TreeNode(``5``);` `        ``root.right.left.left = ``new` `TreeNode(``12``);` `        ``root.right.left.right = ``new` `TreeNode(``7``);` `        ``root.left.right.left = ``new` `TreeNode(``11``);` `        ``root.left.left.right = ``new` `TreeNode(``10``);` `        ``Tree tree = ``new` `Tree(root);` `        ``tree.diagonalSum();` `    ``}` `}`

## Python3

 `# Program to find diagonal sum in a Binary Tree`   `class` `newNode: ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data ` `        ``self``.left ``=` `self``.right ``=` `None` `        `  `# Function to compute height and ` `# root - root of the binary tree ` `# vd - vertical distance diagonally ` `# diagonalSum - map to store Diagonal ` `# Sum(Passed by Reference) ` `def` `diagonalSumUtil(root, vd, diagonalSum) :`   `    ``if``(``not` `root): ` `        ``return` `        `  `    ``if` `vd ``not` `in` `diagonalSum:` `        ``diagonalSum[vd] ``=` `0` `    ``diagonalSum[vd] ``+``=` `root.data `   `    ``# increase the vertical distance` `    ``# if left child ` `    ``diagonalSumUtil(root.left, vd ``+` `1``, ` `                          ``diagonalSum) `   `    ``# vertical distance remains same ` `    ``# for right child ` `    ``diagonalSumUtil(root.right, vd,` `                       ``diagonalSum) `   `# Function to calculate diagonal ` `# sum of given binary tree ` `def` `diagonalSum(root) :`   `    ``# create a map to store Diagonal Sum ` `    ``diagonalSum ``=` `dict``() ` `    `  `    ``diagonalSumUtil(root, ``0``, diagonalSum) `   `    ``print``(``"Diagonal sum in a binary tree is - "``, ` `                                       ``end ``=` `"")` `    `  `    ``for` `it ``in` `diagonalSum:` `        ``print``(diagonalSum[it], end ``=` `" "``)` `        `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``:` `    ``root ``=` `newNode(``1``) ` `    ``root.left ``=` `newNode(``2``) ` `    ``root.right ``=` `newNode(``3``) ` `    ``root.left.left ``=` `newNode(``9``) ` `    ``root.left.right ``=` `newNode(``6``) ` `    ``root.right.left ``=` `newNode(``4``) ` `    ``root.right.right ``=` `newNode(``5``) ` `    ``root.right.left.right ``=` `newNode(``7``) ` `    ``root.right.left.left ``=` `newNode(``12``) ` `    ``root.left.right.left ``=` `newNode(``11``) ` `    ``root.left.left.right ``=` `newNode(``10``) `   `    ``diagonalSum(root)`   `# This code is contributed ` `# by SHUBHAMSINGH10`

Output

`Diagonal sum in a binary tree is - 9 19 42 `

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

### Method 2:

The idea behind this method is inspired by the diagonal relation in matrices. We can observe that all the elements lying on the same diagonal in a matrix have their difference of row and column same. For instance, consider the main diagonal in a square matrix, we can observe the difference of the respective row and column indices of each element on diagonal is same, i.e. each element on the main diagonal have the difference of row and column 0, eg: 0-0, 1-1, 2-2,…n-n.

Similarly, every diagonal has its own unique difference of rows and column, and with the help of this, we can identify each element, that to which diagonal it belongs.

The same idea is applied to solve this problem-

• We will treat the level of the tree nodes as their row indices, and width of the tree nodes as their column indices.
• We will denote the cell of each node as (level, width)

Example- (Taking the same tree as above)

To help you visualize let’s draw a matrix, the first row and first column, are respective width and level indices-

Below is the implementation of the above idea:

## C++

 `// C++ Program to calculate the` `// sum of diagonal nodes.`   `#include ` `using` `namespace` `std;`   `// Node Structure` `struct` `Node {` `    ``int` `data;` `    ``Node *left, *right;` `};`   `// to map the node with level - index` `unordered_map<``int``, ``int``> grid;`   `// Function to create new node` `struct` `Node* newNode(``int` `data)` `{` `    ``struct` `Node* Node` `        ``= (``struct` `Node*)``malloc``(``sizeof``(``struct` `Node));`   `    ``Node->data = data;` `    ``Node->left = Node->right = NULL;`   `    ``return` `Node;` `}`   `// recursvise function to calculate sum of elements` `// where level - index is same.` `void` `addConsideringGrid(Node* root, ``int` `level, ``int` `index)` `{`   `    ``// if there is no child then return` `    ``if` `(root == NULL)` `        ``return``;`   `    ``// add the element in the group of node` `    ``// whose level - index is equal` `    ``grid[level - index] += (root->data);`   `    ``// left child call` `    ``addConsideringGrid(root->left, level + 1, index - 1);`   `    ``// right child call` `    ``addConsideringGrid(root->right, level + 1, index + 1);` `}`   `vector<``int``> diagonalSum(Node* root)` `{` `    ``grid.clear();`   `    ``// Function call` `    ``addConsideringGrid(root, 0, 0);` `    ``vector<``int``> ans;`   `    ``// for different values of level - index` `    ``// add te sum of those node to answer` `    ``for` `(``auto` `x : grid) {` `        ``ans.push_back(x.second);` `    ``}`   `    ``return` `ans;` `}`   `// Driver code` `int` `main()` `{`   `    ``// build binary tree` `    ``struct` `Node* root = newNode(1);` `    ``root->left = newNode(2);` `    ``root->right = newNode(3);` `    ``root->left->left = newNode(9);` `    ``root->left->right = newNode(6);` `    ``root->right->left = newNode(4);` `    ``root->right->right = newNode(5);` `    ``root->right->left->right = newNode(7);` `    ``root->right->left->left = newNode(12);` `    ``root->left->right->left = newNode(11);` `    ``root->left->left->right = newNode(10);`   `    ``// Function Call` `    ``vector<``int``> v = diagonalSum(root);`   `    ``// print the daigonal sums` `    ``for` `(``int` `i = 0; i < v.size(); i++)` `        ``cout << v[i] << ``" "``;` `      ``return` `0;` `}`

Output

`42 9 19 `

Time Complexity-  O(n)

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Improved By : SHUBHAMSINGH10, pchy393

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