<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 parent( TreeNode* root, map<TreeNode*,TreeNode*> &m){
        queue<TreeNode*> q;
        q.push( root );
        while ( !q.empty() ){
            int size = q.size();
            for ( int i = 0; i < size; i++ ){
                TreeNode* front = q.front();
                q.pop();
                if ( front->left != NULL ){
                    m[front->left] = front;
                    q.push( front->left );
                }
                if ( front->right != NULL ){
                    m[front->right] = front;
                    q.push( front->right );
                }
            }
        }
    }
    vector<int> distanceK(TreeNode* root, TreeNode* target, int k) {
        queue<TreeNode*> q;
        map<TreeNode*,TreeNode*> m;
        parent(root,m);
        q.push( target );
        int count = 0;
        map<TreeNode*,bool> visited;
        visited[target] = true;
        while ( !q.empty() ){
            int size = q.size();
            int flag = 0;
            if ( k == count ) break;
            for ( int i = 0; i < size; i++ ){
                TreeNode* front = q.front();
                q.pop();
                if (front->left != NULL && !visited[front->left] ){
                    q.push( front->left);
                    visited[front->left] = true;
                    flag = 1;
                }
                if ( front->right != NULL && !visited[front->right] ){
                    q.push( front->right );
                    visited[front->right] = true;
                    flag = 1;
                }
                if ( m.find(front) != m.end() && !visited[m[front]] ){
                    q.push(m[front]);
                    visited[m[front]] = true;
                    flag = 1;
                }
            }
            count++;
        }
        vector<int> answer;
        while ( !q.empty() ){
            TreeNode* value = q.front();
            int ans = value->val;
            q.pop();
            answer.push_back( ans );
        }
        return answer;
    }
};
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