<p>You are given the <code>root</code> of a binary tree and a positive integer <code>k</code>.</p>

<p>The <strong>level sum</strong> in the tree is the sum of the values of the nodes that are on the <strong>same</strong> level.</p>

<p>Return<em> the </em><code>k<sup>th</sup></code><em> <strong>largest</strong> level sum in the tree (not necessarily distinct)</em>. If there are fewer than <code>k</code> levels in the tree, return <code>-1</code>.</p>

<p><strong>Note</strong> that two nodes are on the same level if they have the same distance from the root.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/12/14/binaryytreeedrawio-2.png" style="width: 301px; height: 284px;" />
<pre>
<strong>Input:</strong> root = [5,8,9,2,1,3,7,4,6], k = 2
<strong>Output:</strong> 13
<strong>Explanation:</strong> The level sums are the following:
- Level 1: 5.
- Level 2: 8 + 9 = 17.
- Level 3: 2 + 1 + 3 + 7 = 13.
- Level 4: 4 + 6 = 10.
The 2<sup>nd</sup> largest level sum is 13.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/12/14/treedrawio-3.png" style="width: 181px; height: 181px;" />
<pre>
<strong>Input:</strong> root = [1,2,null,3], k = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> The largest level sum is 3.
</pre>

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

<ul>
	<li>The number of nodes in the tree is <code>n</code>.</li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= n</code></li>
</ul>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def kthLargestLevelSum(self, root: Optional[TreeNode], k: int) -> int:

        lvl = 1

        if not root:
            return -1

        pq = []

        q = deque([root])

        while q:
            n = len(q)
            tot = 0
            while n:
                n-=1
                node = q.popleft()
                tot += node.val

                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        
            heappush(pq, -tot)

        while len(pq) > 0:
            v = heappop(pq)
            # print(v)
            if lvl == k:
                return -v

            lvl += 1


        return -1
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