# 1080. Insufficient Nodes in Root to Leaf Paths

## Description

Given the root of a binary tree and an integer limit, delete all insufficient nodes in the tree simultaneously, and return the root of the resulting binary tree.

A node is insufficient if every root to leaf path intersecting this node has a sum strictly less than limit.

A leaf is a node with no children.

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]


Constraints:

• The number of nodes in the tree is in the range [1, 5000].
• -105 <= Node.val <= 105
• -109 <= limit <= 109

## Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public TreeNode sufficientSubset(TreeNode root, int limit) {
if (root == null) {
return null;
}
limit -= root.val;
if (root.left == null && root.right == null) {
return limit > 0 ? null : root;
}
root.left = sufficientSubset(root.left, limit);
root.right = sufficientSubset(root.right, limit);
return root.left == null && root.right == null ? null : root;
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* sufficientSubset(TreeNode* root, int limit) {
if (!root) {
return nullptr;
}
limit -= root->val;
if (!root->left && !root->right) {
return limit > 0 ? nullptr : root;
}
root->left = sufficientSubset(root->left, limit);
root->right = sufficientSubset(root->right, limit);
return !root->left && !root->right ? nullptr : root;
}
};

• # 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: Optional[TreeNode], limit: int
) -> Optional[TreeNode]:
if root is None:
return None
limit -= root.val
if root.left is None and root.right is None:
return None if limit > 0 else root
root.left = self.sufficientSubset(root.left, limit)
root.right = self.sufficientSubset(root.right, limit)
return None if root.left is None and root.right is None else root


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func sufficientSubset(root *TreeNode, limit int) *TreeNode {
if root == nil {
return nil
}

limit -= root.Val
if root.Left == nil && root.Right == nil {
if limit > 0 {
return nil
}
return root
}

root.Left = sufficientSubset(root.Left, limit)
root.Right = sufficientSubset(root.Right, limit)

if root.Left == nil && root.Right == nil {
return nil
}
return root
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function sufficientSubset(root: TreeNode | null, limit: number): TreeNode | null {
if (root === null) {
return null;
}
limit -= root.val;
if (root.left === null && root.right === null) {
return limit > 0 ? null : root;
}
root.left = sufficientSubset(root.left, limit);
root.right = sufficientSubset(root.right, limit);
return root.left === null && root.right === null ? null : root;
}


• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} limit
* @return {TreeNode}
*/
var sufficientSubset = function (root, limit) {
if (root === null) {
return null;
}
limit -= root.val;
if (root.left === null && root.right === null) {
return limit > 0 ? null : root;
}
root.left = sufficientSubset(root.left, limit);
root.right = sufficientSubset(root.right, limit);
return root.left === null && root.right === null ? null : root;
};