863. All Nodes Distance K in Binary Tree

2023, May 24 2 mins read

1. Description

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.

2. Example

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: []

3. 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

4. Solutions

Hash Table & Breadth-First Search

n is the number of nodes in root
Time complexity: O(n)
Space complexity: O(n)

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class Solution {
public:
    vector<int> distanceK(TreeNode *root, TreeNode *target, int k) {
        mark_parent(root, nullptr);
        find_results(target, k);

        return result;
    }

private:
    map<TreeNode *, TreeNode *> parent;
    map<TreeNode *, int> is_node_on_path;
    vector<int> result;

    void mark_parent(TreeNode *root, TreeNode *par) {
        if (root != nullptr) {
            parent[root] = par;
            mark_parent(root->left, root);
            mark_parent(root->right, root);
        }
    }

    void find_results(TreeNode *root, int k) {
        if (root != nullptr && is_node_on_path[root] == 0) {
            if (k == 0) {
                result.push_back(root->val);
            } else {
                is_node_on_path[root] = 1;

                find_results(root->left, k - 1);
                find_results(root->right, k - 1);
                find_results(parent[root], k - 1);
            }
        }
    }
};