can we code inorder,preorder and postorder in single code in python ? without using recursion

i want to code all binary tree orders in single code (preorder,postorder,inorder) in O(N) time complexity and O(N) space , using single stack and without recursion.

anybody can help me ?

The below code is easy to keep in mind as we just have to change (3 lines of code), order for left-root-right as we do while using recursion.

Question was asked for Python Implementation, but as Data Structure is language Independent, I am posting this answer so others are benefited.

It is O(N) space, using single stack and without recursion

For Inorder Traversal

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

struct node{
    int val;
    node *left, *right;
    node(int x) {
        val = x;
        left = right = NULL;
    }
};

void traversal_trick(node *root) {
    //inorder
    stack<pair<node*, int>> S;
    
    S.push({root, 0});
    while(!S.empty()){
        pair<node*, int> t = S.top();
        node* cur = t.first;
        int state = t.second;
        
        S.pop();

        if(cur == NULL or state == 3) continue;
        
        S.push({cur, state+1});
        
        if (state == 0) S.push({cur->left, 0});
        else if (state == 1) cout << cur->val << " " ;
        else if (state == 2) S.push({cur->right, 0});
    }
}

int main()
{
    node *root = new node(7); 
    node *t = root;
    root->left = new node(3); root->right = new node(10);
    root->left->left = new node(2); root->left->right = new node(5);
    root->left->left->left = new node(1);
    root->right->left = new node(8); 
    root->right->right = new node(12);
    
    traversal_trick(root);
}

For Preorder Traversal : Just change this part of code

        if (state == 0) cout << cur->val << " " ;
        else if (state == 1) S.push({cur->left, 0});
        else if (state == 2) S.push({cur->right, 0});

For Postorder Traversal : Just change this part of code

        if (state == 0) S.push({cur->left, 0}) ;
        else if (state == 1) S.push({cur->right, 0}) ;
        else if (state == 2) cout << cur->val << " ";

For explanation see this