# openset / leetcode

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## 1080. Insufficient Nodes in Root to Leaf Paths (Medium)

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`

### Hints

Hint 1 Consider a DFS traversal of the tree. You can keep track of the current path sum from root to this node, and you can also use DFS to return the minimum value of any path from this node to the leaf. This will tell you if this node is insufficient.
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