<p>You are given an array of <code>k</code> linked-lists <code>lists</code>, each linked-list is sorted in ascending order.</p>

<p><em>Merge all the linked-lists into one sorted linked-list and return it.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> lists = [[1,4,5],[1,3,4],[2,6]]
<strong>Output:</strong> [1,1,2,3,4,4,5,6]
<strong>Explanation:</strong> The linked-lists are:
[
  1-&gt;4-&gt;5,
  1-&gt;3-&gt;4,
  2-&gt;6
]
merging them into one sorted list:
1-&gt;1-&gt;2-&gt;3-&gt;4-&gt;4-&gt;5-&gt;6
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> lists = []
<strong>Output:</strong> []
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> lists = [[]]
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>k == lists.length</code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= lists[i].length &lt;= 500</code></li>
	<li><code>-10<sup>4</sup> &lt;= lists[i][j] &lt;= 10<sup>4</sup></code></li>
	<li><code>lists[i]</code> is sorted in <strong>ascending order</strong>.</li>
	<li>The sum of <code>lists[i].length</code> will not exceed <code>10<sup>4</sup></code>.</li>
</ul>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class HeapNode:
    def __init__(self,node):
        self.node=node
    def __lt__(self,other):
        return self.node.val < other.node.val
class Solution:
    def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]:
        a = []
        if not lists:
            return 
        if len(lists) == 0:
            return ListNode()
        heapq.heapify(a)
        output=ListNode()
        for l in lists:
            c = l
            while c is not None:
                heappush(a,HeapNode(c))
                c = c.next
                
        print(a)
        if a:
            
            n = ListNode(heappop(a).node.val)
            curr = n
            while len(a) >0:
                curr.next=ListNode(heappop(a).node.val)
                curr = curr.next
            return n
        



            


        
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