1110. Delete Nodes And Return Forest

Problem

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.

Solution (Java)

/**
 * 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 Set<Integer> toDelete;
    private Queue<TreeNode> nodes = new LinkedList<>();

    private TreeNode deleteAndSplit(TreeNode root) {
        if (root == null) {
            return null;
        }
        if (toDelete.contains(root.val)) {
            if (root.left != null) {
                nodes.add(root.left);
            }
            if (root.right != null) {
                nodes.add(root.right);
            }
            return null;
        }
        root.left = deleteAndSplit(root.left);
        root.right = deleteAndSplit(root.right);
        return root;
    }

    public List<TreeNode> delNodes(TreeNode root, int[] localToDelete) {
        toDelete = new HashSet<>();
        for (int node : localToDelete) {
            toDelete.add(node);
        }
        nodes.add(root);
        List<TreeNode> forests = new ArrayList<>();
        while (!nodes.isEmpty()) {
            TreeNode node = nodes.poll();
            node = deleteAndSplit(node);
            if (node != null) {
                forests.add(node);
            }
        }
        return forests;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).