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Problem solution · Java

Delete Nodes And Return Forest

Delete Nodes And Return Forest: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
24 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Delete Nodes And Return Forest, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 24 lines of Java from the credited upstream file 1110.java.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeDelete Nodes And Return Forest · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {    List<TreeNode> ans = new ArrayList<>();    Set<Integer> toDeleteSet = Arrays.stream(to_delete).boxed().collect(Collectors.toSet());    dfs(root, toDeleteSet, true, ans);    return ans;  }   private TreeNode dfs(TreeNode root, Set<Integer> toDeleteSet, boolean isRoot,                       List<TreeNode> ans) {    if (root == null)      return null;     final boolean deleted = toDeleteSet.contains(root.val);    if (isRoot && !deleted)      ans.add(root);     // If root is deleted, both children have the possibility to be a new root    root.left = dfs(root.left, toDeleteSet, deleted, ans);    root.right = dfs(root.right, toDeleteSet, deleted, ans);    return deleted ? null : root;  }}

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