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.
 

DiagonalSum

 

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++

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// C++ Program to find diagonal
// sum in a Binary Tree
#include <bits/stdc++.h>
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

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Java

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// 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<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.
         System.out.print("Diagonal sum in a binary tree is - ");
        while (iterator.hasNext())
        {
            Map.Entry<Integer, Integer> 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();
    }
}

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Python3

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# 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

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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.
This article is contributed by Kumar Gautam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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)

Nodes Level Index Width Index

1

0 0

2



1 -1

3

1 1

4

2 0

5

2 2

6

2 0

7

3 1

9

2 -2

10

3 -1

11

3 -1

12

3 -1

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



  -2

-1

0 1 2
0     1    
1  

2

  3  
2 9   6+4   5
3   10+11+12   7  

Below is the implementation of the above idea:

C++

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// C++ Program to calculate the
// sum of diagonal nodes.
 
#include <bits/stdc++.h>
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;
}

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Output

42 9 19 

Time Complexity-  O(n)

This article is contributed by  Pratik Chaudhary. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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Improved By : SHUBHAMSINGH10, pchy393

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