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
# 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 check(node, pathsum):
            if not node:
                return pathsum >= limit
            pathsum += node.val
            had_children = node.left or node.right
            if not check(node.left, pathsum):
                node.left = None
            if not check(node.right, pathsum):
                node.right = None
            return node.left or node.right if had_children else pathsum >= limit
        
        return root if check(root, 0) else None
# 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 check(node, pathsum):
            if not node:
                return pathsum >= limit
            pathsum += node.val
            had_children = node.left or node.right
            if not check(node.left, pathsum):
                node.left = None
            if not check(node.right, pathsum):
                node.right = None
            return node.left or node.right if had_children else pathsum >= limit
        
        return root if check(root, 0) else None