human
Solve in Python:
Given the root of a binary tree, consider all root to leaf paths: paths from the root to any leaf.  (A leaf is a node with no children.)
A node is insufficient if every such root to leaf path intersecting this node has sum strictly less than limit.
Delete all insufficient nodes simultaneously, and return the root of the resulting binary tree.
 
Example 1:

Input: root = [1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14], limit = 1

Output: [1,2,3,4,null,null,7,8,9,null,14]

Example 2:

Input: root = [5,4,8,11,null,17,4,7,1,null,null,5,3], limit = 22

Output: [5,4,8,11,null,17,4,7,null,null,null,5]
 
Example 3:

Input: root = [1,2,-3,-5,null,4,null], limit = -1

Output: [1,null,-3,4]

 
Note:

The given tree will have between 1 and 5000 nodes.
-10^5 <= node.val <= 10^5
-10^9 <= limit <= 10^9
assistant
#Here is the Python program you requested.

# 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 sufficientSubset(self, root: TreeNode, limit: int) -> TreeNode:
        def helper(root, currsum, limit):
            if not root:
                return True
            currsum += root.val
            if root.left is None and root.right is None:
                return currsum < limit
            
            deleteme = [True, True]
            
            if root.left:
                should_del = helper(root.left, currsum, limit)
                if should_del:
                    del root.left
                    root.left = None
                deleteme[0] = should_del
            
            if root.right:
                should_del = helper(root.right, currsum, limit)
                if should_del:
                    del root.right
                    root.right = None
                deleteme[1] = should_del
            
            return all(deleteme)
        
        deleteroot = helper(root, 0, limit)
        if deleteroot:
            del root
            return None
        return root