*Binary Tree Inorder Traversal

题目：

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?

题解：

中序遍历：递归左 处理当前 递归右。

解法一：递归代码如下：

public void helper(TreeNode root, ArrayList<Integer> re){
        if(root==null)
            return;
        helper(root.left,re);
        re.add(root.val);
        helper(root.right,re);
    }
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        ArrayList<Integer> re = new ArrayList<Integer>();
        if(root==null)
            return re;
        helper(root,re);
        return re;
    }

解法二：非递归解法

public ArrayList<Integer> inorderTraversal(TreeNode root) {  
    ArrayList<Integer> res = new ArrayList<Integer>();  
    if(root == null)  
        return res;  
    LinkedList<TreeNode> stack = new LinkedList<TreeNode>();  
    while(root!=null || !stack.isEmpty()){  
        if(root!=null){
            stack.push(root);
            root = root.left; 
        }else{  
            root = stack.pop();
            res.add(root.val); 
            root = root.right;  
        }  
    }  
    return res;  
}

reference：http://www.cnblogs.com/springfor/p/3877179.html