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TREE-LEVEL-ORDER-TRAVERSAL Solution

/*
*  Author: Arpit Bhayani
*  https://arpitbhayani.me
*/
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <deque>
#include <iostream>
#include <list>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>

#define ll long long

#define MIN(a, b) a < b ? a : b
#define MAX(a, b) a > b ? a : b

using namespace std;

int i = 0;
char ch;
while((ch = getchar()) != '\n') {
str[i++] = ch;
}
str[i] = '\0';
return i;
}

struct node {
int data;
struct node * left;
struct node * right;
};

struct node * new_node(int v) {
struct node * t = (struct node *) calloc(1, sizeof(struct node));
t->data = v;
return t;
}

queue<struct node *> q;
void LevelOrder(struct node * root) {
q.push(root);
while(!q.empty()) {
struct node * p = q.front();
q.pop();

printf("%d ", p->data);

if(p->left) {
q.push(p->left);
}
if(p->right) {
q.push(p->right);
}
}
printf("\n");
}

int main(int argc, char *argv[]) {
struct node * root = new_node(3);
root->left = new_node(5);
root->right = new_node(2);

root->left->left = new_node(1);
root->left->right = new_node(4);

root->right->left = new_node(6);
root->right->right = new_node(7);

root->left->left->right = new_node(9);
root->right->right->left = new_node(8);

LevelOrder(root);

return 0;
}