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].

中序遍历一棵树，我们可以采用递归，也可以用迭代，用迭代的时候借助堆栈来完成。
1，递归

/**
 * 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> list = new ArrayList<Integer>();
        if(root == null) return list;
        getIT(root, list);
        return list;
    }
    public void getIT(TreeNode root, List<Integer> list) {
        if(root == null) return;
        getIT(root.left, list);
        list.add(root.val);
        getIT(root.right, list);
    }
}

2，迭代

/**
 * 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> list = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while(root != null || !stack.isEmpty()) {
            if(root != null) {
                stack.push(root);
                root = root.left;
            } else {
                TreeNode node = stack.pop();
                list.add(node.val);
                root = node.right;
            }
        }
        return list;
    }
}