LeetCode #1080 — MEDIUM

Insufficient Nodes in Root to Leaf Paths

Move from brute-force thinking to an efficient approach using tree strategy.

The Problem

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14]
1

Example 2

[5,4,8,11,null,17,4,7,1,null,null,5,3]
22

Example 3

[1,2,-3,-5,null,4,null]
-1

Core Insight

What unlocks the optimal approach

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.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1080: Insufficient Nodes in Root to Leaf Paths
/**
 * 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;
    }
}

// Accepted solution for LeetCode #1080: Insufficient Nodes in Root to Leaf Paths
/**
 * 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
}

# Accepted solution for LeetCode #1080: Insufficient Nodes in Root to Leaf Paths
# 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

// Accepted solution for LeetCode #1080: Insufficient Nodes in Root to Leaf Paths
struct Solution;
use rustgym_util::*;

trait Postorder {
    fn postorder(self, limit: i32) -> TreeLink;
}

impl Postorder for TreeLink {
    fn postorder(self, limit: i32) -> TreeLink {
        if let Some(node) = self {
            let val = node.borrow_mut().val;
            let left = node.borrow_mut().left.take();
            let right = node.borrow_mut().right.take();
            if left.is_none() && right.is_none() {
                if val >= limit {
                    Some(node)
                } else {
                    None
                }
            } else {
                let l = left.postorder(limit - val);
                let r = right.postorder(limit - val);
                if l.is_some() || r.is_some() {
                    node.borrow_mut().left = l;
                    node.borrow_mut().right = r;
                    Some(node)
                } else {
                    None
                }
            }
        } else {
            None
        }
    }
}

impl Solution {
    fn sufficient_subset(root: TreeLink, limit: i32) -> TreeLink {
        root.postorder(limit)
    }
}

#[test]
fn test() {
    let root = tree!(
        1,
        tree!(
            2,
            tree!(4, tree!(8), tree!(9)),
            tree!(-99, tree!(-99), tree!(-99))
        ),
        tree!(
            3,
            tree!(-99, tree!(12), tree!(13)),
            tree!(7, tree!(-99), tree!(14))
        )
    );
    let limit = 1;
    let res = tree!(
        1,
        tree!(2, tree!(4, tree!(8), tree!(9)), None),
        tree!(3, None, tree!(7, None, tree!(14)))
    );
    assert_eq!(Solution::sufficient_subset(root, limit), res);
    let root = tree!(
        5,
        tree!(4, tree!(11, tree!(7), tree!(1)), None),
        tree!(8, tree!(17), tree!(4, tree!(5), tree!(3)))
    );
    let limit = 22;
    let res = tree!(
        5,
        tree!(4, tree!(11, tree!(7), None), None),
        tree!(8, tree!(17), tree!(4, tree!(5), None))
    );
    assert_eq!(Solution::sufficient_subset(root, limit), res);
    let root = tree!(1, tree!(2, tree!(-5), None), tree!(-3, tree!(4), None));
    let limit = -1;
    let res = tree!(1, None, tree!(-3, tree!(4), None));
    assert_eq!(Solution::sufficient_subset(root, limit), res);
}

// Accepted solution for LeetCode #1080: Insufficient Nodes in Root to Leaf Paths
/**
 * 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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(h)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.