leetcode -- Binary Tree Inorder Traversal

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?

Recursive Version

/**
 * 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) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root == null){
            return result;
        }
        
        if(root.left != null){
            result.addAll(inorderTraversal(root.left));
        }
        result.add(root.val);
        if(root.right != null){
            result.addAll(inorderTraversal(root.right));
        }
        return result;
    }
}

 Iterative Version

public  ArrayList<Integer> inorderTraversal(TreeNode root) {
            // Start typing your Java solution below
            // DO NOT write main() function
            ArrayList<Integer> res = new ArrayList<Integer>();
            if(root==null) return res;
            
            Stack<TreeNode> s = new Stack<TreeNode>();
            TreeNode cur = root;
            while(!s.isEmpty()||cur!=null){
                if(cur!=null){
                    s.push(cur); 
                    cur=cur.left;
                }else{
                    cur=s.pop();
                    res.add(cur.val);
                    cur=cur.right;
                }
            }
            return res;
        }

复杂度：时间O(n)，空间O(logn)

Morris 算法见如下链接

http://gongxuns.blogspot.com/2012/12/leetcodebinary-tree-inorder-traversal.html