Remove all elements from a linked list of integers that have value val.

Example
Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6
Return: 1 --> 2 --> 3 --> 4 --> 5

Solution1：Iterative 遍历 依次比较删除

思路：创建dummy，遍历依次比较删除
实现1_a:没有考虑deleted结点的回收
实现1_b: 考虑deleted结点的回收
Time Complexity: O(N) Space Complexity: O(1)

Solution2：递归(前序) 比较删除

先处理当前level
Time Complexity: O(N) Space Complexity: O(N) 递归缓存

Solution2：递归(后序) 比较删除

后处理当前level
Time Complexity: O(N) Space Complexity: O(N) 递归缓存

Solution1a Code：

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode cur = head, prev = dummy;
        while (cur != null) {
            if (cur.val == val) {
                prev.next = cur.next;
            } else {
                prev = prev.next;
            }
            cur = cur.next;
        }
        return dummy.next;
    }
}

Solution1b Code：

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode cur = head, prev = dummy;
        while (cur != null) {
            ListNode next = cur.next;
            if (cur.val == val) {
                cur.next = null;
                prev.next = next;
                //prev.next = cur.next;
            } else {
                prev = prev.next;
            }
            cur = next;
        }
        return dummy.next;
    }
}

Solution2a Code：

class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if (head == null) return null;
        if (head.val == val) {
            head = removeElements(head.next, val);
        } else {
            head.next = removeElements(head.next, val);
        }
        return head;
    }
}

Solution2b Code：

public ListNode removeElements(ListNode head, int val) {
        if (head == null) return null;
        head.next = removeElements(head.next, val);
        return head.val == val ? head.next : head;
}