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# Insufficient Nodes in Root to Leaf Paths

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:**

1. The given tree will have between `1` and `5000` nodes.
2. `-10^5 <= node.val <= 10^5`
3. `-10^9 <= limit <= 10^9`

```java
class Solution {
    public TreeNode sufficientSubset(TreeNode root, int limit) {
        if (root == null)
            return root;
        boolean ans = helper(root, limit, 0);
        if (ans == true)
            return null;
        return root;
    }

    public boolean helper(TreeNode root, int limit, int sum) {
        if (root.left == null && root.right == null) {
            sum += root.val;
            if (sum < limit)
                return true;
            return false;
        }
        if (root.left != null) {
            boolean ans = helper(root.left, limit, sum + root.val);
            if (ans)
                root.left = null;
        }
        if (root.right != null) {
            boolean ans = helper(root.right, limit, sum + root.val);
            if (ans)
                root.right = null;
        }
        if (root.left == null && root.right == null)
            return true;
        return false;
    }
}
```