1110. Delete Nodes And Return Forest

Description

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.

 

Solutions

Solution: Depth-First Search

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

 

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number[]} to_delete
 * @return {TreeNode[]}
 */
const delNodes = function (root, to_delete) {
  const result = [];

  const dfsTreeNode = node => {
    if (!node) return null;
    const { left, right, val } = node;

    node.left = dfsTreeNode(left);
    node.right = dfsTreeNode(right);

    if (to_delete.includes(val)) {
      node.left && result.push(left);
      node.right && result.push(right);
      return null;
    }
    return node;
  };

  if (!to_delete.includes(root.val)) result.push(root);

  dfsTreeNode(root);
  return result;
};