Iterative Preorder Traversal

Given a Binary Tree, write an iterative function to print Preorder traversal of the given binary tree.

Refer this for recursive preorder traversal of Binary Tree. To convert an inherently recursive procedures to iterative, we need an explicit stack. Following is a simple stack based iterative process to print Preorder traversal.
1) Create an empty stack nodeStack and push root node to stack.
2) Do following while nodeStack is not empty.
….a) Pop an item from stack and print it.
….b) Push right child of popped item to stack
….c) Push left child of popped item to stack

Right child is pushed before left child to make sure that left subtree is processed first.

C++

#include <stdlib.h> 
#include <stdio.h> 
#include <iostream> 
#include <stack> 
  
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; 
} 

Java

// Java program to implement iterative preorder traversal 
import java.util.Stack; 
  
// A binary tree node 
class Node { 
  
    int data; 
    Node left, right; 
  
    Node(int item) { 
        data = item; 
        left = right = null; 
    } 
} 
  
class BinaryTree { 
  
    Node root; 
      
    void iterativePreorder() 
    { 
        iterativePreorder(root); 
    } 
  
    // An iterative process to print preorder traversal of Binary tree 
    void iterativePreorder(Node node) { 
          
        // Base Case 
        if (node == null) { 
            return; 
        } 
  
        // Create an empty stack and push root to it 
        Stack<Node> nodeStack = new Stack<Node>(); 
        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 
            Node mynode = nodeStack.peek(); 
            System.out.print(mynode.data + " "); 
            nodeStack.pop(); 
  
            // Push right and left children of the popped node to stack 
            if (mynode.right != null) { 
                nodeStack.push(mynode.right); 
            } 
            if (mynode.left != null) { 
                nodeStack.push(mynode.left); 
            } 
        } 
    } 
  
    // driver program to test above functions 
    public static void main(String args[]) { 
        BinaryTree tree = new BinaryTree(); 
        tree.root = new Node(10); 
        tree.root.left = new Node(8); 
        tree.root.right = new Node(2); 
        tree.root.left.left = new Node(3); 
        tree.root.left.right = new Node(5); 
        tree.root.right.left = new Node(2); 
        tree.iterativePreorder(); 
  
    } 
} 
  
// This code has been contributed by Mayank Jaiswal 

Python

# Python program to perform iterative preorder traversal 
  
# A binary tree node 
class Node: 
  
    # Constructor to create a new node 
    def __init__(self, data): 
        self.data = data 
        self.left = None
        self.right = None
  
# An iterative process to print preorder traveral of BT 
def iterativePreorder(root): 
      
    # Base CAse  
    if root is None: 
        return 
  
    # create an empty stack and push root to it 
    nodeStack = [] 
    nodeStack.append(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(len(nodeStack) > 0): 
          
        # Pop the top item from stack and print it 
        node = nodeStack.pop() 
        print node.data, 
          
        # Push right and left children of the popped node 
        # to stack 
        if node.right is not None: 
            nodeStack.append(node.right) 
        if node.left is not None: 
            nodeStack.append(node.left) 
      
# Driver program to test above function 
root = Node(10) 
root.left = Node(8) 
root.right = Node(2) 
root.left.left = Node(3) 
root.left.right = Node(5) 
root.right.left = Node(2) 
iterativePreorder(root) 
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007) 

C#

// C# program to implement iterative  
// preorder traversal  
using System; 
using System.Collections.Generic; 
  
// A binary tree node  
public class Node 
{ 
    public int data; 
    public Node left, right; 
  
    public Node(int item) 
    { 
        data = item; 
        left = right = null; 
    } 
} 
  
class GFG 
{ 
public Node root; 
  
public virtual void iterativePreorder() 
{ 
    iterativePreorder(root); 
} 
  
// An iterative process to print preorder  
// traversal of Binary tree  
public virtual void iterativePreorder(Node node) 
{ 
  
    // Base Case  
    if (node == null) 
    { 
        return; 
    } 
  
    // Create an empty stack and push root to it  
    Stack<Node> nodeStack = new Stack<Node>(); 
    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.Count > 0) 
    { 
  
        // Pop the top item from stack and print it  
        Node mynode = nodeStack.Peek(); 
        Console.Write(mynode.data + " "); 
        nodeStack.Pop(); 
  
        // Push right and left children of 
        // the popped node to stack  
        if (mynode.right != null) 
        { 
            nodeStack.Push(mynode.right); 
        } 
        if (mynode.left != null) 
        { 
            nodeStack.Push(mynode.left); 
        } 
    } 
} 
  
// Driver Code  
public static void Main(string[] args) 
{ 
    GFG tree = new GFG(); 
    tree.root = new Node(10); 
    tree.root.left = new Node(8); 
    tree.root.right = new Node(2); 
    tree.root.left.left = new Node(3); 
    tree.root.left.right = new Node(5); 
    tree.root.right.left = new Node(2); 
    tree.iterativePreorder(); 
} 
} 
  
// This code is contributed by Shrikant13 

Output:

10 8 3 5 2 2

This article is compiled by Saurabh Sharma and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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