Iterative Preorder Traversal

Difficulty Level : Easy
Last Updated : 21 Dec, 2020

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++

// 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

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

Time Complexity: O(N)
Auxiliary Space: O(N), where N is the total number of nodes in the tree.

Space Optimized Solution: The idea is to start traversing the tree from root node, and keep printing the left child while exists and simultaneously, push 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.

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

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.

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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