<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:
    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        if(root==NULL) return {};

        map<int,vector<int>>adj;

        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty())
        {
            auto node = q.front();
            q.pop();

            if(node->left)
            {
                adj[node->val].push_back(node->left->val);
                adj[node->left->val].push_back(node->val);
                q.push(node->left);
            }
            if(node->right)
            {
                adj[node->val].push_back(node->right->val);
                adj[node->right->val].push_back(node->val);
                q.push(node->right);
            }
        }
        vector<int>ans;
        queue<pair<int,int>>qq;
        set<int>vis;
        qq.push({target->val,0});
        vis.insert(target->val);
        while(!qq.empty())
        {
            auto front = qq.front();
            qq.pop();

            if(front.second==k)
            {
                ans.push_back(front.first);
                continue;
            }

            for(auto it:adj[front.first])
            {
                if(vis.find(it)==vis.end())
                {
                    qq.push({it,front.second+1});
                    vis.insert(it);
                }
            }
        }
        return ans;

    }
};
15000