<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) {
        
        // this is the process to store the {child, parent} pairs in a hashmap
        map<TreeNode*, TreeNode*> mp;
        queue<TreeNode*> q1;
        q1.push(root);
        while(!q1.empty()) {

            TreeNode *temp = q1.front();
            q1.pop();

            if(temp -> left) {

                mp[temp -> left] = temp;
                q1.push(temp -> left);
            }
            if(temp -> right) {

                mp[temp -> right] = temp;
                q1.push(temp -> right);
            }
        }

        // now we perform BFS
        queue<TreeNode*> q2;
        set<TreeNode*> visited;
        int distance = 0;

        q2.push(target);
        visited.insert(target);

        while(!q2.empty() && distance < k) {

            int n = q2.size();
            for(int i=1; i<=n; i++) {

                TreeNode *temp = q2.front();
                q2.pop();
                
                if(temp -> left && visited.count(temp -> left) == 0) {
                    
                    q2.push(temp -> left);
                    visited.insert(temp -> left);
                }
                if(temp -> right && visited.count(temp -> right) == 0) {
                    
                    q2.push(temp -> right);
                    visited.insert(temp -> right);
                }
                if(mp.count(temp) && visited.count(mp[temp]) == 0) {
                    
                    q2.push(mp[temp]);
                    visited.insert(mp[temp]);
                }
            }
            distance++;
        }

        vector<int> ans;
        while(!q2.empty()) {

            ans.push_back(q2.front() -> val);
            q2.pop();
        }
        return ans;
    }
};
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