Description,
Given a binary tree, return the inorder traversal of its nodes’ values.

For example:

Given binary tree [1,null,2,3],
   1
    \
     2
    /
   3
return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

The leetcode link

Iterative using Stack

This is trivial. Visit the Tree traversal page for detailed guidance.
Tree traversal

Time complexity: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**
    * @solution:       Iterative using stack.
    * @timeComplexity: O(n)
    * @runtime:        1ms
    */
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> ret = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        
        while (!stack.empty() || root != null) {
            if (root != null) {
                stack.push(root);
                root = root.left;
            } else { // pop "in" node of the tree
                root = stack.pop();
                ret.add(root.val);
                root = root.right;
            }
        }
        
        return ret;
    }
}

imgRuntime: 1ms