Is Insufficient Nodes in Root to Leaf Paths easy or hard?

The problem is generally rated Medium difficulty. The traversal itself is straightforward DFS, but the pruning logic requires thinking in post-order so that child nodes are processed before evaluating the parent. Understanding this bottom-up pruning pattern is the main challenge.

Insufficient Nodes in Root to Leaf Paths Python/Java solution

In Python or Java, implement a recursive DFS function that returns the updated subtree after pruning. Track the running path sum and check the limit when reaching leaves. If both left and right children become null after recursion and the path sum is insufficient, return null so the parent removes that node.

How to solve Insufficient Nodes in Root to Leaf Paths in O(n)?

Use a post-order DFS traversal. Pass the current path sum while descending the tree, and when reaching leaves check if the total sum is below the limit. If it is, return null to prune the node. After processing both children, remove the current node if both children become null and the node does not satisfy the limit condition. This ensures each node is processed exactly once.

What is the best approach for Insufficient Nodes in Root to Leaf Paths?

The best approach is a recursive depth-first search that prunes nodes after exploring their children. During DFS, keep track of the cumulative path sum and remove leaf nodes whose path sum is below the limit. When recursion returns, delete any parent node whose children were removed and which no longer leads to a valid path. This runs in O(n) time with O(h) recursion space.

Is Insufficient Nodes in Root to Leaf Paths asked at Google/Amazon/Meta?

Tree pruning and path-sum style problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variations of this problem test understanding of DFS, recursion, and binary tree traversal. Interviewers often expect a clean recursive solution with correct pruning logic.

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

The core data structure is a binary tree. The solution typically uses depth-first search, implemented either with recursion or an explicit stack. The algorithm evaluates root-to-leaf paths and modifies the tree structure by removing insufficient nodes.

What is the time complexity of Insufficient Nodes in Root to Leaf Paths?

The optimal solution runs in O(n) time where n is the number of nodes in the binary tree. Each node is visited once during the DFS traversal and evaluated based on the accumulated path sum. The space complexity is O(h), where h is the tree height due to recursion stack usage.

Description Hints 1 Solution Video 11:11

1080. Insufficient Nodes in Root to Leaf Paths

Medium55.6% acceptance1 company

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

Asked at 1 company

Amazon

Topics

Hints open one at a time

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 maximum value of any path from this node to the leaf. This will tell you if this node is insufficient.

11:11

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