<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://s3-lc-upload.s3.amazonaws.com/uploads/2018/06/28/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>

class Solution {
public:
    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        if (!root) {
            return {};
        }
        findParent(root);
        dfs(target, k);
        return ans;
    }
private:
    std::vector<int> ans;   
    std::unordered_map<int, TreeNode*> parent; // son(n->val) => parent.  
    // Record visited node, store just int val(unique).
    std::unordered_set<int> visit;

    void findParent(TreeNode* node){
        if (!node) {
            return;
        }
        if (node->left) {
            parent[node->left->val] = node;
            findParent(node->left);
        }
        if (node->right) {
            parent[node->right->val] = node;
            findParent(node->right);
        }
    }
    
    void dfs(TreeNode* node, int K) {
        if (!node || visit.find(node->val) != visit.end()) {
            return;
        }
        visit.insert(node->val);
        if (K == 0) {
            ans.push_back(node->val);
            return;
        }
        dfs(node->left, K - 1);
        dfs(node->right, K - 1);
        dfs(parent[node->val], K - 1);
    }
};
16200