iterative preorder traversal

#include <bits/stdc++.h> 
  
using namespace std; 
  
/* A binary tree node has data, left child and right child */
struct node 
{ 
    int data; 
    struct node* left; 
    struct node* right; 
}; 
  
/* Helper function that allocates a new node with the given data and 
   NULL left and right  pointers.*/
struct node* newNode(int data) 
{ 
    struct node* node = new struct node; 
    node->data = data; 
    node->left = NULL; 
    node->right = NULL; 
    return(node); 
} 
  
// An iterative process to print preorder traversal of Binary tree 
void iterativePreorder(node *root) 
{ 
    // Base Case 
    if (root == NULL) 
       return; 
  
    // Create an empty stack and push root to it 
    stack<node *> nodeStack; 
    nodeStack.push(root); 
  
    /* Pop all items one by one. Do following for every popped item 
       a) print it 
       b) push its right child 
       c) push its left child 
    Note that right child is pushed first so that left is processed first */
    while (nodeStack.empty() == false) 
    { 
        // Pop the top item from stack and print it 
        struct node *node = nodeStack.top(); 
        printf ("%d ", node->data); 
        nodeStack.pop(); 
  
        // Push right and left children of the popped node to stack 
        if (node->right) 
            nodeStack.push(node->right); 
        if (node->left) 
            nodeStack.push(node->left); 
    } 
} 
  
// Driver program to test above functions 
int main() 
{ 
    /* Constructed binary tree is 
            10 
          /   \ 
        8      2 
      /  \    / 
    3     5  2 
  */
  struct node *root = newNode(10); 
  root->left        = newNode(8); 
  root->right       = newNode(2); 
  root->left->left  = newNode(3); 
  root->left->right = newNode(5); 
  root->right->left = newNode(2); 
  iterativePreorder(root); 
  return 0; 
} 

// C++ program to print inorder traversal 
// using stack. 
#include<bits/stdc++.h> 
using namespace std; 
  
/* A binary tree Node has data, pointer to left child 
   and a pointer to right child */
struct Node 
{ 
    int data; 
    struct Node* left; 
    struct Node* right; 
    Node (int data) 
    { 
        this->data = data; 
        left = right = NULL; 
    } 
}; 
  
/* Iterative function for inorder tree 
   traversal */
void inOrder(struct Node *root) 
{ 
    stack<Node *> s; 
    Node *curr = root; 
  
    while (curr != NULL || s.empty() == false) 
    { 
        /* Reach the left most Node of the 
           curr Node */
        while (curr !=  NULL) 
        { 
            /* place pointer to a tree node on 
               the stack before traversing 
              the node's left subtree */
            s.push(curr); 
            curr = curr->left; 
        } 
  
        /* Current must be NULL at this point */
        curr = s.top(); 
        s.pop(); 
  
        cout << curr->data << " "; 
  
        /* we have visited the node and its 
           left subtree.  Now, it's right 
           subtree's turn */
        curr = curr->right; 
  
    } /* end of while */
} 
  
/* Driver program to test above functions*/
int main() 
{ 
  
    /* Constructed binary tree is 
              1 
            /   \ 
          2      3 
        /  \ 
      4     5 
    */
    struct Node *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); 
  
    inOrder(root); 
    return 0; 
} 

void iterativePreorder(node *root) 
{ 
    // Base Case 
    if (root == NULL) 
       return; 
  
    // Create an empty stack and push root to it 
    stack<node *> nodeStack; 
    nodeStack.push(root); 
    while (nodeStack.empty() == false) 
    { 
        // Pop the top item from stack and print it 
        struct node *node = nodeStack.top(); 
        printf ("%d ", node->data); 
        nodeStack.pop(); 
  
        // Push right and left children of the popped node to stack 
        if (node->right) 
            nodeStack.push(node->right); 
        if (node->left) 
            nodeStack.push(node->left); 
    } 
}