94. Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

   1
    
     2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

本题要求用迭代来做，如果用迭代来解决，就必须使用stack来解决，里面用来存储下一个需要遍历的node，思路是，创建一个stack和TreeNode，让treenode指向当前的节点，先遍历到最左面的节点，将其值存放在list里面，然后遍历该节点右节点，进入下一步循环（值得注意的是，最左面的节点也可以当作是该节点为中节点），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.isEmpty()){
            while(cur!=null){
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            res.add(cur.val);
            cur = cur.right;
        }
        return res;
    }
}