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BINARY-TREE-ZIGZAG-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 val;
struct node * left;
struct node *right;
};

struct node * new_node(int val) {
struct node *t = (struct node *) malloc(sizeof(struct node));
t->val = val;
t->left = t->right = NULL;
return t;
}

vector< vector<int> > zigzagLevelOrder(struct node * root) {
vector< vector<int> > v;

if(root == NULL) {
return v;
}

int current_level = 0;
queue<struct node *> q;
stack<int> s;

q.push(root);
while(!q.empty()) {
current_level++;
vector<int> level_vector;

// Get from Queue and Put into stack
int x = q.size();
while(x--) {
struct node * t = q.front();
q.pop();
if(current_level % 2 == 0) {
s.push(t->val);
}
else {
level_vector.push_back(t->val);
}

if(t->left) { q.push(t->left); }
if(t->right) { q.push(t->right); }
}

if(current_level % 2 == 0) {
while(!s.empty()) {
level_vector.push_back(s.top());
s.pop();
}
}
v.push_back(level_vector);
}

return v;
}

int main(int argc, char *argv[]) {
struct node * root = NULL;

root = new_node(1);
root->left = new_node(2);
root->right = new_node(3);
root->left->left = new_node(4);
root->right->right = new_node(5);

vector< vector<int> > v = zigzagLevelOrder(root);
for(int i = 0 ; i < v.size(); i++) {
for(int j = 0; j < v[i].size(); j++) {
printf("%d ", v[i][j]);
}
printf("\n");
}

return 0;
}