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?

中文：二叉树的兴许遍历(左-右-根)。能用非递归吗？

递归：

public class BinaryTreePostorderTraversal {

    public List<Integer> postorderTraversal(TreeNode root) {

    	List<Integer> list = new ArrayList<Integer>();

        if(root == null)

        	return list;

        list.addAll(postorderTraversal(root.left));

        list.addAll(postorderTraversal(root.right));

        list.add(root.val);

        return list;

    }

    // Definition for binary tree

    public class TreeNode {

        int val;

        TreeNode left;

        TreeNode right;

        TreeNode(int x) { val = x; }

    }

}

非递归：

    public List<Integer> postorderTraversal(TreeNode root){

    	List<Integer> list = new ArrayList<Integer>();

    	if(root == null)

    		return list;

    	Stack<TreeNode> stack = new Stack<TreeNode>();

    	stack.push(root);//最后訪问

    	while(!stack.isEmpty()){

    		TreeNode current = stack.peek();

    		//根节点无子节点

    		if(current.left == null && current.right == null){

    			list.add(current.val);

    			stack.pop();

    		}

    		if(current.left != null){

    			stack.push(current.left);

    			current.left = null;

    			continue;

    		}

    		if(current.right != null){

    			stack.push(current.right);

    			current.right = null;

    			continue;

    		}

    	}

    	return list;

    }