<p>Given the <code>root</code> of a binary tree and an integer <code>limit</code>, delete all <strong>insufficient nodes</strong> in the tree simultaneously, and return <em>the root of the resulting binary tree</em>.</p>

<p>A node is <strong>insufficient</strong> if every root to <strong>leaf</strong> path intersecting this node has a sum strictly less than <code>limit</code>.</p>

<p>A <strong>leaf</strong> is a node with no children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/06/05/insufficient-11.png" style="width: 500px; height: 207px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14], limit = 1
<strong>Output:</strong> [1,2,3,4,null,null,7,8,9,null,14]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/06/05/insufficient-3.png" style="width: 400px; height: 274px;" />
<pre>
<strong>Input:</strong> root = [5,4,8,11,null,17,4,7,1,null,null,5,3], limit = 22
<strong>Output:</strong> [5,4,8,11,null,17,4,7,null,null,null,5]
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/06/11/screen-shot-2019-06-11-at-83301-pm.png" style="width: 250px; height: 199px;" />
<pre>
<strong>Input:</strong> root = [1,2,-3,-5,null,4,null], limit = -1
<strong>Output:</strong> [1,null,-3,4]
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 5000]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= limit &lt;= 10<sup>9</sup></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 helper(self, node, curr_val, limit):
        if node.left == None and node.right == None:
            if curr_val + node.val < limit:
                return 0
            else:
                return 1
        
        if node.left:
            left = self.helper(node.left, curr_val + node.val, limit)
        else:
            left = 0
        
        if node.right:
            right = self.helper(node.right, curr_val + node.val, limit)
        else:
            right = 0

        if not left:
            node.left = None
        if not right:
            node.right = None

        if not left and not right:
            return 0
        else:
            return 1

    def sufficientSubset(self, root: Optional[TreeNode], limit: int) -> Optional[TreeNode]:
        exists = self.helper(root, 0, limit)
        if exists: 
            return root
        else:
            return None
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