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?

原题链接：https://oj.leetcode.com/problems/binary-tree-inorder-traversal/

题目：给定一二叉树，返回其节点值的中序遍历。

中序遍历：左子-》根-》右子

递归解法：

	public List<Integer> inorderTraversal(TreeNode root) {
		List<Integer> list = new ArrayList<Integer>();
		if(root == null)
			return list;
		list.addAll(inorderTraversal(root.left));
		list.add(root.val);
		list.addAll(inorderTraversal(root.right));
		return list;
	}

	// Definition for binary tree
	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}

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