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We're a place where coders share, stay up-to-date and grow their careers. # Inorder Traversal - Iterative Rohith V
Full Stack Engineer || Java || Graduated From Government Engineerig College Thrissur. Interested in Data Structures in Java

## Problem Statement

Given a binary tree. Find the inorder traversal of the tree without using recursion.

## Sample Test Cases

Example 1

Input:

``````           1
/    \
2       3
/   \
4     5
``````

Output: 4 2 5 1 3
Explanation:
Inorder traversal (Left->Root->Right) of
the tree is 4 2 5 1 3.

Example 2

Input:

``````            8
/   \
1      5
\     /  \
7   10   6
\  /
10 6
``````

Output: 1 7 10 8 6 10 5 6
Explanation:
Inorder traversal (Left->Root->Right)
of the tree is 1 7 10 8 6 10 5 6.

You don't need to read input or print anything. Your task is to complete the function inOrder() which takes the root of the tree as input and returns a list containing the inorder traversal of the tree, calculated without using recursion.

Expected time complexity: O(N)
Expected auxiliary space: O(N)

Constraints:
1 <= Number of nodes <= 10^5
1 <= Data of a node <= 10^5

## What is inorder traversal

It is a type of traversal in tree where first we go to left of current node, then the node, then to the right of the node
That is left -> root -> right.

It can be easily done using recursion:

``````ArrayList<Integer> inOrder(Node root)
{
// Code
if (root == null)
return new ArrayList<>();
ArrayList<Integer> result = new ArrayList<>();
inorder(root, result);
return result;
}

void inorder(Node node, ArrayList<Integer> result) {
if (node == null)
return;
inorder(node.left, result);
inorder(node.right, result);
}
``````

It can also be done using iterative method where we use a stack and do the processing

### Iterative Approach

So here we make use of stack as usual and move to the left of the root until we reach the leaf.
Then that left node is added to the result and processed back and check with the right node.

• Let us have a current node = root. `current = root`.

• We also make use of a stack . `Stack<Node> stack = new Stack<>()`.

• Until the stack is null or current is null, we start the outer loop

• Now inside this loop, we move to the left of current node until it is null and each time we add the nodes into the stack.

``````while (current != null) {
stack.push(current)l
current = current.left;
}
``````
• Now if the current reaches null, it means we reach the leaf node, so we pop it out, add it into the result and then check with current.right
``````current = stack.pop();
current = current.right;
``````
• Finally return the result.

## Code walkthrough

``````       1
/    \
2       3
/   \
4     5
``````
• stack = [] result = [] current = 1

• push 1 -> stack result = [] current.left = 2

• push 2 -> stack[2, 1] result = [] current.left = 4

• push 4 -> stack[4, 2, 1] result = [] current.left = null

• pop 4 -> stack[2, 1] result =  current = 4

• right of 4 is also null current.right = null

• pop 2 -> stack result = [4, 2] current.right = 5

• push 5 -> stack[5, 1] result = [4, 2] current.left = null

• pop 5 -> stack result = [4, 2, 5] we processed with left child of 1 current = 1

• pop 1 -> stack[] result = [4, 2, 5, 1] current goes to right of 1, ie 3 current.right = 3

• push 3 -> stack result = [4, 2, 5, 1] current.left = null

• pop 3 -> stack[] result = [4, 2, 5, 1, 3] current.right = null

Now both stack is empty and current is null, so we return result

## Complexity

time complexity: O(N)
auxiliary space: O(N)

## Code

``````class Solution
{
// Return a list containing the inorder traversal of the given tree
ArrayList<Integer> inOrder(Node root)
{
// Code
if (root == null)
return new ArrayList<>();
ArrayList<Integer> result = new ArrayList<>();
Stack<Node> stack = new Stack<>();
Node current = root;
while (!stack.isEmpty() || current != null) {
while (current != null) {
stack.push(current);
current = current.left;
}
current = stack.pop();