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?

Morris遍歷參考：Morris

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {   
    //使用線索二叉樹，Morris遍歷
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root==null) return res;
        TreeNode n = root;
        TreeNode pre = null;
        while(n!=null){
            if(n.left==null){//當前節點的左節點為空，則列印當前節點，並設定當前節點為當前節點的右節點
                res.add(n.val);
                n = n.right;
            }else{//否則，找到當前節點的中序遍歷前驅節點，即當前節點左孩子的最右側孩子
                pre = n.left;
                while(pre.right!=null&&pre.right!=n) pre = pre.right;
                if(pre.right == null){//如果前驅節點為空，則設定其右節點為當前節點（設定線索），並將當前節點設為當前節點的左孩子
                    pre.right = n;
                    n = n.left;
                }else{//否則，說明當前節點的左分支已經遍歷過，恢復成原樣，列印，並設定當前節點為當前節點的右孩子
                    pre.right = null;
                    res.add(n.val);
                    n = n.right;
                }
            }
            
        }
        return res;       
    }
}