LeetCode #1110 — MEDIUM

Delete Nodes And Return Forest

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

Solve on LeetCode

The Problem

Problem Statement

Given the root of a binary tree, each node in the tree has a distinct value.

After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees).

Return the roots of the trees in the remaining forest. You may return the result in any order.

Example 1:

Input: root = [1,2,3,4,5,6,7], to_delete = [3,5]
Output: [[1,2,null,4],[6],[7]]

Example 2:

Input: root = [1,2,4,null,3], to_delete = [3]
Output: [[1,2,4]]

Constraints:

The number of nodes in the given tree is at most 1000.
Each node has a distinct value between 1 and 1000.
to_delete.length <= 1000
to_delete contains distinct values between 1 and 1000.

Step 01

Brute Force Baseline

Problem summary: Given the root of a binary tree, each node in the tree has a distinct value. After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees). Return the roots of the trees in the remaining forest. You may return the result in any order.

Baseline thinking

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

Pattern signal: Array · Hash Map · Tree

Example 1

[1,2,3,4,5,6,7]
[3,5]

Example 2

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

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #1110: Delete Nodes And Return Forest
/**
 * 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 {
    private boolean[] s = new boolean[1001];
    private List<TreeNode> ans = new ArrayList<>();

    public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
        for (int x : to_delete) {
            s[x] = true;
        }
        if (dfs(root) != null) {
            ans.add(root);
        }
        return ans;
    }

    private TreeNode dfs(TreeNode root) {
        if (root == null) {
            return null;
        }
        root.left = dfs(root.left);
        root.right = dfs(root.right);
        if (!s[root.val]) {
            return root;
        }
        if (root.left != null) {
            ans.add(root.left);
        }
        if (root.right != null) {
            ans.add(root.right);
        }
        return null;
    }
}

// Accepted solution for LeetCode #1110: Delete Nodes And Return Forest
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func delNodes(root *TreeNode, to_delete []int) (ans []*TreeNode) {
	s := make([]bool, 1001)
	for _, x := range to_delete {
		s[x] = true
	}
	var dfs func(*TreeNode) *TreeNode
	dfs = func(root *TreeNode) *TreeNode {
		if root == nil {
			return nil
		}
		root.Left = dfs(root.Left)
		root.Right = dfs(root.Right)
		if !s[root.Val] {
			return root
		}
		if root.Left != nil {
			ans = append(ans, root.Left)
		}
		if root.Right != nil {
			ans = append(ans, root.Right)
		}
		return nil
	}
	if dfs(root) != nil {
		ans = append(ans, root)
	}
	return
}

# Accepted solution for LeetCode #1110: Delete Nodes And Return Forest
# 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 delNodes(
        self, root: Optional[TreeNode], to_delete: List[int]
    ) -> List[TreeNode]:
        def dfs(root: Optional[TreeNode]) -> Optional[TreeNode]:
            if root is None:
                return None
            root.left, root.right = dfs(root.left), dfs(root.right)
            if root.val not in s:
                return root
            if root.left:
                ans.append(root.left)
            if root.right:
                ans.append(root.right)
            return None

        s = set(to_delete)
        ans = []
        if dfs(root):
            ans.append(root)
        return ans

// Accepted solution for LeetCode #1110: Delete Nodes And Return Forest
struct Solution;
use rustgym_util::*;
use std::collections::HashSet;
use std::iter::FromIterator;

trait Postorder {
    fn postorder(self, nodes: &HashSet<i32>) -> (TreeLink, Vec<TreeLink>);
}

impl Postorder for TreeLink {
    fn postorder(self, nodes: &HashSet<i32>) -> (TreeLink, Vec<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();
            let (left_root, left_forest) = left.postorder(nodes);
            let (right_root, right_forest) = right.postorder(nodes);
            let mut forest = vec![];
            for t in left_forest {
                forest.push(t);
            }
            for t in right_forest {
                forest.push(t);
            }
            if nodes.contains(&val) {
                if left_root.is_some() {
                    forest.push(left_root);
                }
                if right_root.is_some() {
                    forest.push(right_root);
                }
                (None, forest)
            } else {
                node.borrow_mut().left = left_root;
                node.borrow_mut().right = right_root;
                (Some(node), forest)
            }
        } else {
            (None, vec![])
        }
    }
}

impl Solution {
    fn del_nodes(root: TreeLink, to_delete: Vec<i32>) -> Vec<TreeLink> {
        let nodes = HashSet::from_iter(to_delete);
        let mut res = vec![];
        let (root, forest) = root.postorder(&nodes);
        if root.is_some() {
            res.push(root);
        }
        for t in forest {
            res.push(t);
        }
        res
    }
}

#[test]
fn test() {
    let root = tree!(
        1,
        tree!(2, tree!(4), tree!(5)),
        tree!(3, tree!(6), tree!(7))
    );
    let to_delete = vec![3, 5];
    let res = vec![tree!(1, tree!(2, tree!(4), None), None), tree!(6), tree!(7)];
    assert_eq!(Solution::del_nodes(root, to_delete), res);
}

// Accepted solution for LeetCode #1110: Delete Nodes And Return Forest
/**
 * 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 delNodes(root: TreeNode | null, to_delete: number[]): Array<TreeNode | null> {
    const s: boolean[] = Array(1001).fill(false);
    for (const x of to_delete) {
        s[x] = true;
    }
    const ans: Array<TreeNode | null> = [];
    const dfs = (root: TreeNode | null): TreeNode | null => {
        if (!root) {
            return null;
        }
        root.left = dfs(root.left);
        root.right = dfs(root.right);
        if (!s[root.val]) {
            return root;
        }
        if (root.left) {
            ans.push(root.left);
        }
        if (root.right) {
            ans.push(root.right);
        }
        return null;
    };
    if (dfs(root)) {
        ans.push(root);
    }
    return ans;
}

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(n)

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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.