Iterative Preorder Traversal

Given a Binary Tree, write an iterative function to print the Preorder traversal of the given binary tree.
Refer to this for recursive preorder traversal of Binary Tree. To convert an inherently recursive procedure 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 the following while nodeStack is not empty. 
   1. Pop an item from the stack and print it. 
   2. Push right child of a popped item to stack 
   3. Push left child of a popped item to stack

The right child is pushed before the left child to make sure that the left subtree is processed first.

C++

// C++ program to implement 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;
}

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

Python3

# 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 traversal 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, end=" ")

        # 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

Javascript

<script>

// Javascript program to implement iterative
// preorder traversal

// A binary tree node
class Node
{
    constructor(item)
    {
        this.data = item;
        this.left = null;
        this.right = null;
    }
}

var root = null;

// An iterative process to print preorder
// traversal of Binary tree
function iterativePreorder(node)
{

    // Base Case
    if (node == null)
    {
        return;
    }

    // Create an empty stack and push root to it
    var 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.length > 0)
    {

        // Pop the top item from stack and print it
        var mynode = nodeStack[nodeStack.length - 1];
        document.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
root = new Node(10);
root.left = new Node(8);
root.right = new Node(2);
root.left.left = new Node(3);
root.left.right = new Node(5);
root.right.left = new Node(2);

iterativePreorder(root);

// This code is contributed by itsok

</script>

Output

10 8 3 5 2 2 

Time Complexity: O(N) 
Auxiliary Space: O(H), where H is the height of the tree.

Another Solution: In the previous solution we can see that the left child is popped as soon as it is pushed to the stack, therefore it is not required to push it into the stack. 

The idea is to start traversing the tree from the root node, and keep printing the left child while exists and simultaneously, push the right child of every node in an auxiliary stack. Once we reach a null node, pop a right child from the auxiliary stack and repeat the process while the auxiliary stack is not-empty. 

This is a micro-optimization over the previous approach, both the solutions use asymptotically similar auxiliary space.

Below is the implementation of the above approach: 

C++

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

// Tree Node
struct Node {
    int data;
    Node *left, *right;

    Node(int data)
    {
        this->data = data;
        this->left = this->right = NULL;
    }
};

// Iterative function to do Preorder traversal of the tree
void preorderIterative(Node* root)
{
    if (root == NULL)
        return;

    stack<Node*> st;

    // start from root node (set current node to root node)
    Node* curr = root;

    // run till stack is not empty or current is
    // not NULL
    while (!st.empty() || curr != NULL) {
        // Print left children while exist
        // and keep pushing right into the
        // stack.
        while (curr != NULL) {
            cout << curr->data << " ";

            if (curr->right)
                st.push(curr->right);

            curr = curr->left;
        }

        // We reach when curr is NULL, so We
        // take out a right child from stack
        if (st.empty() == false) {
            curr = st.top();
            st.pop();
        }
    }
}

// Driver Code
int main()
{
    Node* root = new Node(10);
    root->left = new Node(20);
    root->right = new Node(30);
    root->left->left = new Node(40);
    root->left->left->left = new Node(70);
    root->left->right = new Node(50);
    root->right->left = new Node(60);
    root->left->left->right = new Node(80);

    preorderIterative(root);

    return 0;
}

Java

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 preorderIterative()
{
    preorderIterative(root);
}

// Iterative function to do Preorder
// traversal of the tree
void preorderIterative(Node node)
{
    if (node == null)
    {
        return;
    }

    Stack<Node> st = new Stack<Node>();

    // Start from root node (set curr
    // node to root node)
    Node curr = node;

    // Run till stack is not empty or
    // current is not NULL
    while (curr != null || !st.isEmpty())
    {

        // Print left children while exist
        // and keep pushing right into the
        // stack.
        while (curr != null)
        {
            System.out.print(curr.data + " ");

            if (curr.right != null)
                st.push(curr.right);

            curr = curr.left;
        }

        // We reach when curr is NULL, so We
        // take out a right child from stack
        if (!st.isEmpty())
        {
            curr = st.pop();
        }
    }
}

// Driver code
public static void main(String args[])
{
    BinaryTree tree = new BinaryTree();

    tree.root = new Node(10);
    tree.root.left = new Node(20);
    tree.root.right = new Node(30);
    tree.root.left.left = new Node(40);
    tree.root.left.left.left = new Node(70);
    tree.root.left.right = new Node(50);
    tree.root.right.left = new Node(60);
    tree.root.left.left.right = new Node(80);

    tree.preorderIterative();
}
}

// This code is contributed by Vivek Singh Bhadauria

Python3

# Tree Node
class Node:

    def __init__(self, data = 0):
        self.data = data
        self.left = None
        self.right = None

# Iterative function to do Preorder traversal of the tree
def preorderIterative(root):

    if (root == None):
        return

    st = []

    # start from root node (set current node to root node)
    curr = root

    # run till stack is not empty or current is
    # not NULL
    while (len(st) or curr != None):

        # Print left children while exist
        # and keep appending right into the
        # stack.
        while (curr != None):

            print(curr.data, end = " ")

            if (curr.right != None):
                st.append(curr.right)

            curr = curr.left

        # We reach when curr is NULL, so We
        # take out a right child from stack
        if (len(st) > 0):
            curr = st[-1]
            st.pop()

# Driver Code

root = Node(10)
root.left = Node(20)
root.right = Node(30)
root.left.left = Node(40)
root.left.left.left = Node(70)
root.left.right = Node(50)
root.right.left = Node(60)
root.left.left.right = Node(80)

preorderIterative(root)

# This code is contributed by Arnab Kundu

C#

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;
    }
}

public class BinaryTree{

Node root;

void preorderIterative()
{
    preorderIterative(root);
}

// Iterative function to do Preorder
// traversal of the tree
void preorderIterative(Node node)
{
    if (node == null)
    {
        return;
    }

    Stack<Node> st = new Stack<Node>();

    // Start from root node (set curr
    // node to root node)
    Node curr = node;

    // Run till stack is not empty or
    // current is not NULL
    while (curr != null || st.Count!=0)
    {

        // Print left children while exist
        // and keep pushing right into the
        // stack.
        while (curr != null)
        {
            Console.Write(curr.data + " ");

            if (curr.right != null)
                st.Push(curr.right);

            curr = curr.left;
        }

        // We reach when curr is NULL, so We
        // take out a right child from stack
        if (st.Count != 0)
        {
            curr = st.Pop();
        }
    }
}

// Driver code
public static void Main(String []args)
{
    BinaryTree tree = new BinaryTree();

    tree.root = new Node(10);
    tree.root.left = new Node(20);
    tree.root.right = new Node(30);
    tree.root.left.left = new Node(40);
    tree.root.left.left.left = new Node(70);
    tree.root.left.right = new Node(50);
    tree.root.right.left = new Node(60);
    tree.root.left.left.right = new Node(80);

    tree.preorderIterative();
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

class Node
{
    constructor(item)
    {
        this.left = null;
        this.right = null;
        this.data = item;
    }
}

let root;

// Iterative function to do Preorder
// traversal of the tree
function preorderiterative(node)
{
    if (node == null)
    {
        return;
    }

    let st = [];

    // Start from root node (set curr
    // node to root node)
    let curr = node;

    // Run till stack is not empty or
    // current is not NULL
    while (curr != null || st.length > 0)
    {

        // Print left children while exist
        // and keep pushing right into the
        // stack.
        while (curr != null)
        {
            document.write(curr.data + " ");

            if (curr.right != null)
                st.push(curr.right);

            curr = curr.left;
        }

        // We reach when curr is NULL, so We
        // take out a right child from stack
        if (st.length > 0)
        {
            curr = st.pop();
        }
    }
}

function preorderIterative()
{
    preorderiterative(root);
}

// Driver code
root = new Node(10);
root.left = new Node(20);
root.right = new Node(30);
root.left.left = new Node(40);
root.left.left.left = new Node(70);
root.left.right = new Node(50);
root.right.left = new Node(60);
root.left.left.right = new Node(80);

preorderIterative();

// This code is contributed by decode2207

</script>

Output

10 20 40 70 80 50 30 60 

Time Complexity: O(N) 
Auxiliary Space: O(H), where H is the height of the tree.

ANOTHER APPROACH:

Intuition:

Using Morris Traversal, we can traverse the tree without using stack and recursion. The algorithm for Preorder is almost similar to Morris traversal for Inorder.

1…If left child is null, print the current node data. Move to right child. 
….Else, Make the right child of the inorder predecessor point to the current node. Two cases arise: 
………a) The right child of the inorder predecessor already points to the current node. Set right child to NULL. Move to right child of current node. 
………b) The right child is NULL. Set it to the current node. Print the current node’s data and move to left child of current node. 
2…Iterate until the current node is not NULL.

Implementation:

Java

// Java program to implement Morris preorder traversal

// A binary tree node
class Node {

    int data;
    Node left, right;

    Node(int item)
    {
        data = item;
        left = right = null;
    }
}

class BinaryTree {

    Node root;

    void morrisTraversalPreorder()
    {
        morrisTraversalPreorder(root);
    }

    // Preorder traversal without recursion and without
    // stack
    void morrisTraversalPreorder(Node node)
    {
        while (node != null) {

            // If left child is null, print the current node
            // data. Move to right child.
            if (node.left == null) {
                System.out.print(node.data + " ");
                node = node.right;
            }
            else {

                // Find inorder predecessor
                Node current = node.left;
                while (current.right != null
                       && current.right != node) {
                    current = current.right;
                }

                // If the right child of inorder predecessor
                // already points to this node
                if (current.right == node) {
                    current.right = null;
                    node = node.right;
                }

                // If right child doesn't point to this
                // node, then print this node and make right
                // child point to this node
                else {
                    System.out.print(node.data + " ");
                    current.right = node;
                    node = node.left;
                }
            }
        }
    }

    void preorder() { preorder(root); }

    // Function for Standard preorder traversal
    void preorder(Node node)
    {
        if (node != null) {
            System.out.print(node.data + " ");
            preorder(node.left);
            preorder(node.right);
        }
    }

    // Driver programs to test above functions
    public static void main(String args[])
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
        tree.root.right.left = new Node(6);
        tree.root.right.right = new Node(7);
        tree.root.left.left.left = new Node(8);
        tree.root.left.left.right = new Node(9);
        tree.root.left.right.left = new Node(10);
        tree.root.left.right.right = new Node(11);
        tree.morrisTraversalPreorder();
        System.out.println("");
        tree.preorder();
    }
}

// this code has been contributed by Raunak Singh

Output

1 2 4 8 9 5 10 11 3 6 7 
1 2 4 8 9 5 10 11 3 6 7 

Time Complexity: O(n), we visit every node at most once.
Auxiliary Space: O(1), we use a constant amount of space for variables and pointers.