1. Digonal Traverasal.cpp

// https://www.geeksforgeeks.org/problems/diagonal-traversal-of-binary-tree/1

/*
Logic :
1. Start with root and traverse for size of queue , under the while(!Q.emtpy())
2. Put left of nodes in queue and datat of node in ans vector and keep moving right , and do this for size of that queus
*/

// GFG

#include <bits/stdc++.h>
using namespace std;

struct Node
{
    int data;
    Node *left, *right;
};

Node *newNode(int data)
{
    Node *node = new Node;
    node->data = data;
    node->left = node->right = NULL;
    return node;
}

vector<vector<int>> result;
void diagonalPrint(Node *root)
{
    if (root == NULL)
        return;

    queue<Node *> q;
    q.push(root);

    while (!q.empty())
    {
        int size = q.size();
        vector<int> answer;

        while (size--)
        {
            Node *temp = q.front();
            q.pop();

            // traversing each component;
            while (temp)
            {
                answer.push_back(temp->data);

                if (temp->left)
                    q.push(temp->left);

                temp = temp->right;
            }
        }
        result.push_back(answer);
    }
}

int main()
{
    Node *root = newNode(8);
    root->left = newNode(3);
    root->right = newNode(10);
    root->left->left = newNode(1);
    root->left->right = newNode(6);
    root->right->right = newNode(14);
    root->right->right->left = newNode(13);
    root->left->right->left = newNode(4);
    root->left->right->right = newNode(7);

    diagonalPrint(root);

    for (int i = 0; i < result.size(); i++)
    {
        for (int j = 0; j < result[i].size(); j++)
            cout << result[i][j] << " ";
        cout << endl;
    }

    return 0;
}