<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>

/**
 * 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:
    void solve(TreeNode*root,vector<TreeNode*>adj[]){
        if(!root) return;
        if(root->left) {
            adj[root->val].push_back(root->left);
            adj[root->left->val].push_back(root);
        }
       if(root->right) {
            adj[root->val].push_back(root->right);
            adj[root->right->val].push_back(root);
        }
        solve(root->left,adj);
        solve(root->right,adj);
    }
    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        vector<TreeNode*>adj[502];
        solve(root,adj);
        int vis[502]={0};
        queue<pair<TreeNode*,int>>q;
        vector<int>ans;
        q.push({target,0});
        vis[target->val]=1;
        while(!q.empty()){
            auto[node,d]=q.front();
            q.pop();
            if(d==k) ans.push_back(node->val);
            for(auto &x:adj[node->val]){
                if(!vis[x->val]){
                    vis[x->val]=1;
                    q.push({x,d+1});
                }
            }
        }
        return ans;
    }
};
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