# Partition a linked list

Given the head of a linked list and a value x, partition it such that all nodes **less than** x come before nodes **greater than or equal** to x.

You should **preserve** the original relative order of the nodes in each of the two partitions.

```
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode d1 = new ListNode();
ListNode d2 = new ListNode();
ListNode t1 = d1, t2 = d2;
while (head != null) {
if (head.val < x) {
t1.next = head;
t1 = t1.next;
} else {
t2.next = head;
t2 = t2.next;
}
head = head.next;
}
t1.next = d2.next;
t2.next = null;
return d1.next;
}
}
```

## Related Problems

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.

Return *the head of the merged linked list*.

Given the root of a binary tree, flatten the tree into a "linked list":

The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.

The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should **not** change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a **unique answer**.

Return *the root of the trimmed binary search tree*. Note that the root may change depending on the given bounds.

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