Question

我尝试自己动手进行迭代后序遍历。我的解决方案在Leetcode Online Judge

中超出时间限制

public List<Integer> postorderTraversalIterativel(TreeNode root) {
    List<Integer> ret = new LinkedList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();
    TreeNode cur = root;
    while (cur != null || !stack.isEmpty()) {
        if (cur != null) {
            stack.push(cur);
            cur = cur.left;
        } else {
            TreeNode parent = stack.peek(), child = null;
            if (parent.right == null) {
                // pop hard
                stack.pop();
                while (parent.right == child && !stack.isEmpty()) {
                    child = parent;
                    ret.add(child.val);
                    parent = stack.pop();
                }
            } else {
                cur = parent.right;
            }
        }
    }
    return ret;
}

虽然wikipedia的官方实施可以通过测试。

public List<Integer> postorderTraversalIterativel(TreeNode root) {
    List<Integer> ret = new LinkedList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();
    TreeNode cur = root, lastVisited = null;
    while (cur != null || !stack.isEmpty()) {
        if (cur != null) {
            stack.push(cur);
            cur = cur.left;
        } else {
            TreeNode parent = stack.peek();
            if (parent.right != null && lastVisited != parent.right) {
                // make sure pop all the node has right child of the
                // previous pop
                cur = parent.right;
            } else {
                stack.pop();
                ret.add(parent.val);
                lastVisited = parent;
            }
        }
    }
    return ret;
}

通过检查代码，我无法理解为什么我的实现比官方实现慢。谁能指出发生了什么？（可能我的解决方案在逻辑上是错误的，但是我的解决方案失败了，我已经进行了单元测试，单元测试很快就完成了......）。欢迎提出任何意见。

public void testPostorderTraversal1() {
    TreeNode root = new TreeNode(3);
    TreeNode right = new TreeNode(1);
    TreeNode rightLeft = new TreeNode(2);
    root.right = right;
    right.left = rightLeft;

    List<Integer> list = new LinkedList<Integer>();
    list.add(2);
    list.add(1);
    list.add(3);

    assertEquals(list.toString(), sut.postorderTraversal(root).toString());
}

public void testPostorderTraversal2() {
    TreeNode root = new TreeNode(1);
    TreeNode right = new TreeNode(2);
    root.right = right;

    List<Integer> list = new LinkedList<Integer>();
    list.add(2);
    list.add(1);

    assertEquals(list.toString(), sut.postorderTraversal(root).toString());
}

Answer 1

您的代码可以执行无限循环。

你的'testPostorderTraversal1'unitest没有完成。

如果有任何节点具有已离开子节点的右子节点，则无法完成。

Tree Postorder Traversal表现

1 个答案: