Iterative Preorder Traversal
- Difficulty Level : Easy
- Last Updated : 17 Jun, 2022
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.
![]()
- Create an empty stack nodeStack and push root node to stack.
- Do the following while nodeStack is not empty.
- Pop an item from the stack and print it.
- Push right child of a popped item to stack
- 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()
(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
):
(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.My Personal Notes arrow_drop_upRecommended ArticlesPage :Article Contributed By :Improved By :Article Tags :Practice Tags :