<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 func(TreeNode* node,int& k, vector<int>& ans,unordered_map<int,bool>& visited,unordered_map<TreeNode*,TreeNode*>& parent,int dist){
        if(node==NULL || visited[node->val]) return;

        visited[node->val] = true;
        if(dist==k) ans.push_back(node->val);

        if(node->left) func(node->left,k,ans,visited,parent,dist+1);
        if(node->right) func(node->right,k,ans,visited,parent,dist+1);

        if(parent[node]) func(parent[node],k,ans,visited,parent,dist+1);

    }
    void findParent(TreeNode* root, unordered_map<TreeNode*,TreeNode*>& parent){

        if(root==NULL) return;

        if(root->left){
            parent[root->left] = root;
            findParent(root->left,parent);
        }

        if(root->right){
            parent[root->right] = root;
            findParent(root->right,parent);
        }

    }


    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {

        unordered_map<int,bool> visited;
        // unordered_map<int,int> m;

        unordered_map<TreeNode*,TreeNode*> parent;

        parent[root]=NULL;

        findParent(root,parent);

        vector<int> ans;
        func(target,k,ans,visited,parent,0);
        return ans;  
    }
};
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