<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) {
        unordered_map<TreeNode*, TreeNode*> parent_track; // node -> parent
        unordered_map<TreeNode*, bool> visited;
        queue<TreeNode*> queue;
        queue.push(root);
        while(!queue.empty()) { /*First BFS to get a track of parent nodes*/
            TreeNode* current = queue.front(); queue.pop();
            if(current->left) {
                parent_track[current->left] = current;
                queue.push(current->left);
            }
            if(current->right) {
                parent_track[current->right] = current;
                queue.push(current->right);
            }
        }
        queue.push(target);
        visited[target] = true;
        int curr_level = 0;
        while(!queue.empty()) { /*Second BFS to go upto K level from target node and using our hashtable info*/
            int size = queue.size();
            if(curr_level++ == K) break;
            for(int i=0; i<size; i++) {
                TreeNode* current = queue.front(); queue.pop();
                if(current->left && !visited[current->left]) {
                    queue.push(current->left);
                    visited[current->left] = true;
                }
                if(current->right && !visited[current->right]) {
                    queue.push(current->right);
                    visited[current->right] = true;
                }
                if(parent_track[current] && !visited[parent_track[current]]) {
                    queue.push(parent_track[current]);
                    visited[parent_track[current]] = true;
                }
            }
        }
        vector<int> result;
        while(!queue.empty()) {
            TreeNode* current = queue.front(); queue.pop();
            result.push_back(current->val);
        }
        return result;
    }
};
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