The correct, optimal and working solution for programming question binary-tree-inorder-traversal on leetcode

/* * 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 readline(char *str) { 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; }