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

For example:
Given binary tree {1,#,2,3},
return [3,2,1].

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

題意：后序遍歷，即左右中。

思路：
1、分治遞歸遍歷左右中。

public List<Integer> postorderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    if (root == null) {
        return res;
    }

    helper(root, res);

    return res;
}

private void helper(TreeNode node, List<Integer> res) {
    if (node == null) {
        return;
    }

    helper(node.left, res);
    helper(node.right, res);
    res.add(node.val);
}

2、修改前序遍歷為中右左，再把list翻轉(zhuǎn)就變成了左右中。

public List<Integer> postorderTraversal1(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    if (root == null) {
        return res;
    }

    Stack<TreeNode> stack = new Stack<>();
    stack.push(root);
    while (!stack.isEmpty()) {
        TreeNode cur = stack.pop();
        res.add(cur.val);
        if (cur.left != null) {
            stack.push(cur.left);
        }
        if (cur.right != null) {
            stack.push(cur.right);
        }
    }

    Collections.reverse(res);

    return res;
}