863. All Nodes Distance K in Binary Tree

Problem

Given the root of a binary tree, the value of a target node target, and an integer k, return **an array of the values of all nodes that have a distance *k* from the target node.**

You can return the answer in any order.

  Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, k = 2
Output: [7,4,1]
Explanation: The nodes that are a distance 2 from the target node (with value 5) have values 7, 4, and 1.

Example 2:

Input: root = [1], target = 1, k = 3
Output: []

  Constraints:

• The number of nodes in the tree is in the range [1, 500].

• 0 <= Node.val <= 500

• All the values Node.val are unique.

• target is the value of one of the nodes in the tree.

• 0 <= k <= 1000

Solution (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private void kFar(TreeNode root, int k, TreeNode visited, List<Integer> ls) {
        if (root == null || k < 0 || root == visited) {
            return;
        }
        if (k == 0) {
            ls.add(root.val);
            return;
        }
        kFar(root.left, k - 1, visited, ls);
        kFar(root.right, k - 1, visited, ls);
    }

    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        List<Integer> ls = new ArrayList<>();
        if (k == 0) {
            ls.add(target.val);
            return ls;
        }
        nodeToRoot(root, target, k, ls);
        return ls;
    }

    private int nodeToRoot(TreeNode root, TreeNode target, int k, List<Integer> ls) {
        if (root == null) {
            return -1;
        }
        if (root.val == target.val) {
            kFar(root, k, null, ls);
            return 1;
        }
        int ld = nodeToRoot(root.left, target, k, ls);
        if (ld != -1) {
            kFar(root, k - ld, root.left, ls);
            return ld + 1;
        }
        int rd = nodeToRoot(root.right, target, k, ls);
        if (rd != -1) {
            kFar(root, k - rd, root.right, ls);
            return rd + 1;
        }
        return -1;
    }
}

Explain:

nope.

Complexity:

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