Problem Statement
Medium
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 <= 1000to_delete contains distinct values between 1 and 1000.
/**
* 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 {
List<TreeNode> forest = new ArrayList<>();
private TreeNode dfs(TreeNode root, Set<Integer> del){
if(root==null)
return null;
root.left = dfs(root.left, del);
root.right = dfs(root.right, del);
if(del.contains(root.val)){
if(root.left!=null)
forest.add(root.left);
if(root.right!=null)
forest.add(root.right);
root = null;
}
return root;
}
public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
Set<Integer> del = Arrays.stream(to_delete).boxed().collect(Collectors.toSet());
root = dfs(root, del);
if(root!=null)
forest.add(root);
return forest;
}
}