How do you solve Insufficient Nodes in Root to Leaf Paths on LeetCode?

Insufficient Nodes in Root to Leaf Paths (LeetCode #1080) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: Understanding the difference between leaf nodes and internal nodes is crucial for this problem.

What is the brute force solution for Insufficient Nodes in Root to Leaf Paths?

The brute force approach for Insufficient Nodes in Root to Leaf Paths is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we can traverse the tree and calculate the sum of each root-to-leaf path. If any path sum is less than the limit, we mark the nodes along that path for deletion. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Insufficient Nodes in Root to Leaf Paths?

Insufficient Nodes in Root to Leaf Paths uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Tree Traversal.

What are the interview tips for Insufficient Nodes in Root to Leaf Paths?

Always clarify the problem requirements and constraints before jumping into coding. Think about edge cases, such as trees with only one node or negative values. Practice explaining your thought process while coding, as communication is key in interviews.

What are common mistakes when solving Insufficient Nodes in Root to Leaf Paths?

Failing to correctly handle the case where a node has only one child. Not considering the case where the root itself might be insufficient.

#1080

Insufficient Nodes in Root to Leaf Paths

Medium

TreeDepth-First SearchBinary TreeDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

💡

Intuition

Time O(n)Space O(h)

The optimal approach uses a depth-first search (DFS) to calculate the path sum while traversing the tree. Instead of checking all paths after the fact, we can determine if a node is insufficient as we go, allowing us to prune the tree efficiently.

⚙️

Algorithm

3 steps

1Step 1: Perform a DFS traversal of the tree, keeping track of the current path sum.
2Step 2: If a leaf node is reached, check if the path sum is less than the limit.
3Step 3: If a node is insufficient, remove it by returning null for that node.

solution.py25 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def sufficientSubset(root, limit):
9    if not root:
10        return None
11    def dfs(node, current_sum):
12        if not node:
13            return current_sum < limit
14        current_sum += node.val
15        if not node.left and not node.right:
16            return current_sum < limit
17        left_insufficient = dfs(node.left, current_sum)
18        right_insufficient = dfs(node.right, current_sum)
19        if left_insufficient:
20            node.left = None
21        if right_insufficient:
22            node.right = None
23        return not node.left and not node.right
24    dfs(root, 0)
25    return root if root.val >= limit else None

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1Understanding the difference between leaf nodes and internal nodes is crucial for this problem.
2Using DFS allows us to efficiently prune the tree while calculating path sums.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.