Approach: Recursion from the right

def removeNodes(self, head):
    if not head:
        return None
    head.next = self.removeNodes(head.next)
    if head.next and head.next.val > head.val:
        return head.next
    return head

Complexity: O(n) time, O(n) space.

Full Solution — all languages

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def removeNodes(self, head):
        if not head:
            return None
        head.next = self.removeNodes(head.next)
        if head.next and head.next.val > head.val:
            return head.next
        return head

import java.util.*;
class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}
class Solution {
    public ListNode removeNodes(ListNode head) {
        if (head == null) return null;
        head.next = removeNodes(head.next);
        if (head.next != null && head.next.val > head.val) return head.next;
        return head;
    }
}

#include <bits/stdc++.h> using namespace std;
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x): val(x), next(nullptr) {
    }
};
class Solution {
    public: ListNode* removeNodes(ListNode* head) {
        if (!head) return nullptr;
        head->next = removeNodes(head->next);
        if (head->next && head->next->val > head->val) return head->next;
        return head;
    }
};

/** Node shape: { val, next }. Build/return plain objects or use this ctor. */ function ListNode(val, next) {
    this.val = (val === undefined ? 0 : val);
    this.next = (next === undefined ? null : next);
}
/**
 * @param {ListNode} head
 * @return {ListNode}
 */ var removeNodes = function(head) {
    if (!head) return null;
    head.next = removeNodes(head.next);
    if (head.next && head.next.val > head.val) return head.next;
    return head;
};

Remove Nodes From Linked List

Example 1

Example 2

Constraints