#### Inorder Traversal

S
m
```package com.thealgorithms.datastructures.trees;

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;

/**
* Given tree is traversed in an 'inorder' way: LEFT -> ROOT -> RIGHT.
* Below are given the recursive and iterative implementations.
*
* Complexities:
* Recursive: O(n) - time, O(n) - space, where 'n' is the number of nodes in a tree.
*
* Iterative: O(n) - time, O(h) - space, where 'n' is the number of nodes in a tree
* and 'h' is the height of a binary tree.
* In the worst case 'h' can be O(n) if tree is completely unbalanced, for instance:
* 5
*  \
*   6
*    \
*     7
*      \
*       8
*
* @author Albina Gimaletdinova on 21/02/2023
*/
public class InorderTraversal {
public static List<Integer> recursiveInorder(BinaryTree.Node root) {
List<Integer> result = new ArrayList<>();
recursiveInorder(root, result);
return result;
}

public static List<Integer> iterativeInorder(BinaryTree.Node root) {
List<Integer> result = new ArrayList<>();
if (root == null) return result;

Deque<BinaryTree.Node> stack = new ArrayDeque<>();
while (!stack.isEmpty() || root != null) {
while (root != null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
root = root.right;
}
return result;
}

private static void recursiveInorder(BinaryTree.Node root, List<Integer> result) {
if (root == null) {
return;
}
recursiveInorder(root.left, result);