**Linked List Deletion - Amazon Top Interview Questions**

### Problem Statement :

Given a singly linked list node, and an integer target, return the same linked list with all nodes whose value is target removed. Constraints n ≤ 100,000 where n is the number of nodes in node Example 1 Input node = [0, 1, 1, 2] target = 1 Output [0, 2]

### Solution :

` ````
Solution in C++ :
** Recursive
LLNode* solve(LLNode* node, int target) {
if(not node) return node;
node->next = solve(node->next, target);
return node->val == target ? node->next : node;
}
** Iterative
LLNode* solve(LLNode* head, int target) {
auto* node = head;
while (node) {
while (node->next && node->next->val == target) {
node->next = node->next->next;
}
node = node->next;
}
return (head->val == target ? head->next : head);
}
```

` ````
Solution in Java :
import java.util.*;
/**
* class LLNode {
* int val;
* LLNode next;
* }
*/
class Solution {
public LLNode solve(LLNode node, int target) {
LLNode prev = node;
LLNode cur = node.next;
while (cur != null) {
LLNode nxt = cur.next;
if (cur.val == target)
prev.next = nxt;
else {
prev = cur;
}
cur = nxt;
}
if (node.val == target)
return node.next;
return node;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, node, target):
while node != None and node.val == target:
node = node.next
ret = node
while node != None:
while node.next != None and node.next.val == target:
node.next = node.next.next
node = node.next
return ret
```

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