Binary Tree Inorder Traversal leetcode

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

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

Inorder: return [1,3,2]. 左中右

addAll 是因为arraylist result 里面应该添加int

第一种是递归的算法，时间复杂度是O(n), 而空间复杂度则是递归栈的大小，即O(logn)。

第二种是迭代的算法， 用栈来实现。 时间复杂度是O(n), 空间O(logn)
第三种是分治算法

public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        helper(res, root);
        return res;
    }
    public void helper(List<Integer> res, TreeNode root) {
        if (root == null) {
            return;
        }
        helper(res, root.left);
        res.add(root.val);
        helper(res, root.right);
    }
}
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while (!stack.isEmpty() || root != null) {
            if (root != null) {
                stack.push(root);
                root= root.left;
            } else {
                root = stack.pop();
                res.add(root.val);
                root = root.right;
            }
        }
        return res;
    }
}
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        List<Integer> left = inorderTraversal(root.left);
        List<Integer> right = inorderTraversal(root.right);
        res.addAll(left);
        res.add(root.val);
        res.addAll(right);
        return res;    
    }
    
    
}