Introduction

In this blog post, we will explore how to solve the LeetCode problem "1080" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's an efficient solution to the problem of deleting insufficient nodes in a binary tree:

• Recursive Depth-First Search (DFS): Traverse the tree using DFS, passing the current path sum down to each node.
• Pruning Insufficient Nodes: If a node's value added to the path sum results in a path sum less than the limit for all paths leading from the root, remove that node.

• Post-Order Traversal with Modification: We perform pruning in a post-order fashion, allowing us to determine the insufficient nature of parent nodes based on their children's status.

• Time Complexity: O(N), where N is the number of nodes in the tree.

• Space Complexity: O(H), where H is the height of the tree, due to the recursive call stack. In the worst case (skewed tree), H = N; in the best case (balanced tree), H = log N.

Code

    class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def delete_insufficient_nodes(root, limit):
    """
    Deletes insufficient nodes in a binary tree.

    Args:
        root: The root of the binary tree.
        limit: The integer limit.

    Returns:
        The root of the resulting binary tree.
    """

    def is_insufficient(node, current_sum):
        if not node:
            return True

        current_sum += node.val

        if not node.left and not node.right:  # Leaf node
            return current_sum < limit

        left_insufficient = is_insufficient(node.left, current_sum)
        right_insufficient = is_insufficient(node.right, current_sum)

        if left_insufficient:
            node.left = None
        if right_insufficient:
            node.right = None

        return left_insufficient and right_insufficient

    if is_insufficient(root, 0):
        return None
    return root