# BINARY-TREE-INORDER-TRAVERSAL leetcode Solution - Correct, Optimal and Working

``````/*
*  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<int> in_order(struct node * root) {
vector<int> order;
stack<struct node *> s;
struct node * p = root;

int is_done = 0;

while(is_done == 0) {
if(p != NULL) {
s.push(p);
p = p->left;
}
else {
if(!s.empty()) {
struct node * t = s.top();
s.pop();
order.push_back(t->val);
p = t->right;
}
else {
is_done = 1;
}
}
}
return order;
}

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->left->right = new_node(5);
root->right->left = new_node(6);

vector<int> order = in_order(root);
for(vector<int>::iterator itr = order.begin(); itr != order.end(); itr++) {
printf("%d ", *itr);
}
printf("\n");

return 0;
}
``````