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?

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在做99. Recover Binary Search Tree 时，要求不用O（n)的空间复杂度，只用constant space，因此不能用普通的迭代和递归法，递归里有函数栈，迭代里有用户声明的栈，都会带来Ｏ（lgn)-O(n)的空间复杂度。

用线索二叉树。

在迭代中，栈里存的都是根节点，是为了遍历完左子树后能重新找到根，再遍历根和右子树。想一种办法，能不用栈又能找回根节点。办法是：对每个根节点，首先找到在左子树里的它的前继节点，前继节点的right指针是闲置的，把闲置的right指针指向根，就可以重新找到根了。

     什么时候恢复ｒｉｇｈｔ指针、并且在遍历这个根节点的时候如何知道，已经遍历过这个根节点的左子树，还是要开始遍历整个根下的树？

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        TreeNode cur = root, prev = null;
        List<Integer> re = new LinkedList<Integer>();
        while (cur != null) {
            prev = cur.left;
            if (prev == null) {
                re.add(cur.val); // leaf node, traverse current node
                cur = cur.right; // jump to its next
                continue;
            }
            while (true) {
                if (prev.right == null) {
                    prev.right = cur;
                    cur = cur.left;
                    break;
                } else if (prev.right == cur) {
                    prev.right = null;
                    re.add(cur.val); // root node, traverse current node
                    cur = cur.right;
                    break;
                }
                prev = prev.right;
            }
        }
        return re;
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> re = new LinkedList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode tmp = root, right;
        while (tmp != null) {
            stack.push(tmp);
            tmp = tmp.left;
        }
        while (!stack.isEmpty()) {
            tmp = stack.pop();
            re.add(tmp.val);
            right = tmp.right;
            while (right != null) {
                stack.push(right);
                right = right.left;
            }
        }
        return re;
    }
}