Java [Leetcode 94]Binary Tree Inorder Traversal

题目描述：

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].

解题思路：

使用栈。从根节点开始迭代循环访问，将节点入栈，并循环将左子树入栈。如果当前节点为空，则弹出栈顶节点，也就是当前节点的父节点，并将父节点的值加入到list中，然后选择右节点作为循环的节点，依次循环。

代码如下：

/**
 * 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> res = new ArrayList<Integer> ();
    	Stack<TreeNode> stack = new Stack<TreeNode> ();
    	TreeNode cur = root;
    	while(cur != null || !stack.empty()){
    		while(cur != null){
    			stack.push(cur);
    			cur = cur.left;
    		}
    		cur =stack.pop();
    		res.add(cur.val);
    		cur = cur.right;
    	}
    	return res;
    }
}