Skip to content

1080. Insufficient Nodes in Root to Leaf Paths 👎

  • Time: $O(n)$
  • Space: $O(h)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
class Solution {
 public:
  TreeNode* sufficientSubset(TreeNode* root, int limit) {
    if (root == nullptr)
      return nullptr;
    if (root->left == nullptr && root->right == nullptr)
      return root->val < limit ? nullptr : root;
    root->left = sufficientSubset(root->left, limit - root->val);
    root->right = sufficientSubset(root->right, limit - root->val);
    return root->left == nullptr && root->right == nullptr ? nullptr : root;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution {
  public TreeNode sufficientSubset(TreeNode root, int limit) {
    if (root == null)
      return null;
    if (root.left == null && root.right == null)
      return root.val < limit ? null : root;
    root.left = sufficientSubset(root.left, limit - root.val);
    root.right = sufficientSubset(root.right, limit - root.val);
    return root.left == null && root.right == null ? null : root;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
class Solution:
  def sufficientSubset(
      self,
      root: TreeNode | None,
      limit: int
  ) -> TreeNode | None:
    if not root:
      return None
    if not root.left and not root.right:
      return None if root.val < limit else root
    root.left = self.sufficientSubset(root.left, limit - root.val)
    root.right = self.sufficientSubset(root.right, limit - root.val)
    return None if not root.left and not root.right else root