[2018-Interview] Binary Tree Inorder Traversal

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

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

My answer: iterative solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        if (root == null) {
            return result;
        }
        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
        Set<TreeNode> visitedNodes = new HashSet<TreeNode>();
        stack.addLast(root);

        while(!stack.isEmpty()) {
            TreeNode current = stack.peekLast();
            if (current.left != null && !visitedNodes.contains(current.left)) {
                stack.addLast(current.left);
                continue;
            } 
            
                result.add(current.val);
                visitedNodes.add(current);
                stack.removeLast();
                if (current.right != null) {
                    stack.addLast(current.right);
                }
            
        }
        
        return result;
    }
}

Jiuzhang solution:

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Inorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        ArrayList<Integer> result = new ArrayList<Integer>();
        TreeNode curt = root;
        while (curt != null || !stack.empty()) {
            while (curt != null) {
                stack.add(curt);
                curt = curt.left;
            }
            curt = stack.pop();
            result.add(curt.val);
            curt = curt.right;
        }
        return result;
    }
}