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

    For example:
    Given binary tree {1,#,2,3},
       1
        \
         2
        /
       3
    return [1,3,2].
    
    Note: Recursive solution is trivial, could you do it iteratively?
 */
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> res = new ArrayList<Integer>();
        Stack<TreeNode> s = new Stack<TreeNode>();
        TreeNode curr = root;
        boolean done = false;
        while(!done){
            if(curr != null){
                s.push(curr);
                curr = curr.left;
            }else{
                if(s.isEmpty())
                    done = true;
                else{
                    curr = s.pop();
                    res.add(curr.val);
                    curr = curr.right;
                }
            }
        }
        return res;
    }
}