94. Binary Tree Inorder Traversal

题目：
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?

解答：

1. Recursive:

public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<Integer>();
    if (root == null) return result;
    if (root.left == null && root.right == null) {
        result.add(root.val);
        return result;
    }
    
    //合并两个list,直接addAll就好啦
    result.addAll(inorderTraversal(root.left));
    result.add(root.val);
    result.addAll(inorderTraversal(root.right));
    
    return result;
}

1. Iteratively

//Iterative的方法很巧妙，当时想了很久也没做出来，所以这里标注一下
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
        TreeNode curt = root;
        while (curt != null || !stack.isEmpty()) {
            while (curt != null) {
                stack.push(curt);
                curt = curt.left;
            }
            curt = stack.pop();
            list.add(curt.val);
            curt = curt.right;
        }
        return list;
    }