All Nodes Distance K in Binary Tree
<p>Given the <code>root</code> of a binary tree, the value of a target node <code>target</code>, and an integer <code>k</code>, return <em>an array of the values of all nodes that have a distance </em><code>k</code><em> from the target node.</em></p>

<p>You can return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0800-0899/0863.All%20Nodes%20Distance%20K%20in%20Binary%20Tree/images/sketch0.png" style="width: 500px; height: 429px;" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, k = 2
<strong>Output:</strong> [7,4,1]
Explanation: The nodes that are a distance 2 from the target node (with value 5) have values 7, 4, and 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1], target = 1, k = 3
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 500]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 500</code></li>
	<li>All the values <code>Node.val</code> are <strong>unique</strong>.</li>
	<li><code>target</code> is the value of one of the nodes in the tree.</li>
	<li><code>0 &lt;= k &lt;= 1000</code></li>
</ul>
Tree; Depth-First Search; Breadth-First Search; Hash Table; Binary Tree
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        unordered_map<TreeNode*, TreeNode*> g;
        vector<int> ans;

        auto dfs = [&](this auto&& dfs, TreeNode* node, TreeNode* fa) {
            if (!node) return;
            g[node] = fa;
            dfs(node->left, node);
            dfs(node->right, node);
        };

        auto dfs2 = [&](this auto&& dfs2, TreeNode* node, TreeNode* fa, int k) {
            if (!node) return;
            if (k == 0) {
                ans.push_back(node->val);
                return;
            }
            for (auto&& nxt : {node->left, node->right, g[node]}) {
                if (nxt != fa) {
                    dfs2(nxt, node, k - 1);
                }
            }
        };

        dfs(root, nullptr);
        dfs2(target, nullptr, k);
        return ans;
    }
};
