# Diagonal Sum of a Binary Tree in C++?

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To consider the nodes that are passing between lines of slope -1. The diagonal sum of the binary tree will be calculated by the sum of all nodes data that are present between these lines of reference.

Let us first define the struct that would represent a tree node that contains the data and its left and right node child. If this is the first node to be created then it’s a root node otherwise a child node.

struct Node {
int data;
struct Node *leftChild, *rightChild;
};

Next we create our createNode(int data) function that takes an int value and assign it to the data member of the node after creating the new node. The function returns the pointer to the created node.

Node * createNode(int data){
Node * node = new Node;
node->data = data;
return node;
}

Next, the diagonal_sum(Node *root, int depth, map<int, int> &diagonalSum) takes the root node, current depth and the diagonalSum map by reference. If root isn’t NULL then we add the current root data to the current depth index in the diagonalSum map to get the sum of elements .We then perform an recursive inorder traversal on the tree and add 1 to the depth whenever we move to the left child.

void diagonal_sum(Node *root, int depth, map<int, int> &diagonalSum){
if(root){
diagonalSum[depth]+=root->data;
diagonal_sum(root->leftChild, depth+1, diagonalSum);
diagonal_sum(root->rightChild, depth, diagonalSum);
}
}

Inside our main function we create a tree using the createNode(data) method and also create a map SumMap. The root node , current depth which is 1 and the sumMap are sent to diagonal_sum where the sumMap is filled with key-values pair. Next we create our iterator it for iterating over the sumMap map.

int main(){
Node *root = createNode(1);
root->rightChild = createNode(3);
root->rightChild->leftChild = createNode(4);
root->rightChild->leftChild->leftChild = createNode(12);
root->rightChild->leftChild->rightChild = createNode(7);
root->leftChild = createNode(2);
root->leftChild->leftChild = createNode(9);
root->leftChild->rightChild = createNode(6);
root->leftChild->leftChild->rightChild = createNode(10);
root->leftChild->rightChild->leftChild = createNode(11);
root->rightChild->rightChild = createNode(5);
map<int,int> sumMap;
diagonal_sum(root, 1, sumMap);
map<int,int>::iterator it;

Finally we iterate on our SumMap by using the iterator it inside our for loop and print the diagonal sums.

for(it=sumMap.begin(); it!=sumMap.end();++it){
int value = it->second;
cout<<value<<"\t";
}

## Example

Let us see the following implementation for finding the diagonal sum of a binary tree.

Live Demo

#include<iostream>
#include<map>
using namespace std;
struct Node{
int data;
struct Node* leftChild, *rightChild;
};
Node * createNode(int data){
Node * node = new Node;
node->data = data;
return node;
}
void diagonal_sum(Node *root, int depth, map<int, int> &diagonalSum){
if(root){
diagonalSum[depth]+=root->data;
diagonal_sum(root->leftChild, depth+1, diagonalSum);
diagonal_sum(root->rightChild, depth, diagonalSum);
}
}
int main(){
Node *root = createNode(1);
root->rightChild = createNode(3);
root->rightChild->leftChild = createNode(4);
root->rightChild->leftChild->leftChild = createNode(12);
root->rightChild->leftChild->rightChild = createNode(7);
root->leftChild = createNode(2);
root->leftChild->leftChild = createNode(9);
root->leftChild->rightChild = createNode(6);
root->leftChild->leftChild->rightChild = createNode(10);
root->leftChild->rightChild->leftChild = createNode(11);
root->rightChild->rightChild = createNode(5);
map<int,int> sumMap;
diagonal_sum(root, 1, sumMap);
map<int,int>::iterator it;
for(it=sumMap.begin(); it!=sumMap.end();++it){
int value = it->second;
cout<<value<<"\t";
}
return 0;
}

## Output

The above code will produce the following output −

91942