145. Binary Tree Postorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [3,2,1].

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

本题其实和Binary Tree Inorder Traversal比较类似的，只不过是相反的，本题不可以按照正常的顺序来做，因为左子树然后右子树是比较难以转换的，因此可以采用倒着来，即先当前节点，然后是其右子树，做法和之前的中序是一样的，只不过是相反的思路，最后遍历完，把链表list反转过来，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> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        TreeNode node = root;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while(!stack.isEmpty()||node!=null){
            if(node!=null){
                stack.push(node);
                res.add(node.val);
                node = node.right;
            }else{
                node = stack.pop().left;
            }
        }
        Collections.reverse(res);
        return res;
    }
}