Friday, September 4, 2020

LeetCode [1110] Delete Nodes And Return Forest

 1110. Delete Nodes And Return Forest

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]]

 

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.
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/**
 * 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 {
    Set<Integer> deleted;
    List<TreeNode> ret;
    public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
        deleted = new HashSet<>();
        ret = new ArrayList<>();
        for(int i : to_delete){
            deleted.add(i);
        }
        helper(root, true);
        return ret;
    }
    
    TreeNode helper(TreeNode node, boolean isRoot){
        if(node==null){
            return null;
        }
        
        boolean isDeleted = deleted.contains(node.val);
        if(isRoot && !isDeleted){
            ret.add(node);
        }
        
        node.left = helper(node.left, isDeleted);
        node.right = helper(node.right, isDeleted);
        
        return isDeleted ? null : node;
    }
}
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