<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:
TreeNode* findtarget(TreeNode*target,TreeNode*root){
    if(root == nullptr) return nullptr;

    if(root->val == target->val) return root;
    TreeNode*left = findtarget(target,root->left);
    TreeNode*right = findtarget(target,root->right);

    if(!left && !right) return nullptr;
    if(!left) return right;
    return left;
}
void findparent(TreeNode*root,TreeNode*parent,unordered_map<TreeNode*,TreeNode*> &mp){
    if(root == nullptr) return;

    mp[root] = parent;
    findparent(root->left,root,mp);
    findparent(root->right,root,mp);
}
    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        vector<int> ans;
        if(root == nullptr) return ans;
        TreeNode*tar = findtarget(target,root);
        unordered_map<TreeNode*,TreeNode*> mp;
        findparent(root,nullptr,mp);
        unordered_map<TreeNode*,int> vis;
        vis[tar] = 1;
        queue<TreeNode*> q;
        q.push(tar);
        int level = 0;
        while(!q.empty()){
            int size = q.size();
            while(size--){
            TreeNode*node = q.front();
            q.pop();
            if(level == k){
             ans.push_back(node->val);
            }
            if(node->left && vis[node->left]!=1){
                q.push(node->left);
                vis[node->left] = 1;
            }
            if(node->right && vis[node->right]!=1){
                q.push(node->right);
                vis[node->right] = 1;
            }
            if(mp[node]!=nullptr && vis[mp[node]]!=1){
                q.push(mp[node]);
                vis[mp[node]] = 1;
            }
            }
            level++;
        }
        return ans;
    }
};
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