<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) 
{
    queue<TreeNode*>q;

    // child,parent
    map<TreeNode*,TreeNode*>parent;
    q.push(root);

    // root has no parent
    while(!q.empty())
    {
        int size=q.size();
        for(int i=0;i<size;i++)
        {
            TreeNode *node = q.front();
            q.pop();

            if(node->left)
            {
                parent[node->left] = node;
                q.push(node->left);
            }
            if(node->right)
            {
                parent[node->right] = node;
                q.push(node->right);
            }
        }
    }
    //---------------------------------------
    // q is now empty and can be reused

    map<TreeNode*,bool>vis;
    // start from target, and apply bfs (just like graph)
    q.push(target);
    vis[target]=true;

    int dist=0;
    while(!q.empty())
    {
        if(dist==k)
            break;

        int size=q.size();
        for(int i=0;i<size;i++)
        {
            TreeNode*node = q.front();
            q.pop();

            if(node->left and vis[node->left]==false)
            {
                q.push(node->left);
                vis[node->left]=true;
            }
            if(node->right  and vis[node->right]==false)
            {
                q.push(node->right);
                vis[node->right]=true;
            }
            if(parent.find(node)!=parent.end()  and vis[parent[node]] ==false)
            {
                q.push(parent[node]);
                vis[parent[node]]=true;
            }

        }

        dist+=1;
    }

    vector<int>res;
    while(!q.empty())
    {
        res.push_back(q.front()->val);
        q.pop();
    }

    return res;
}
};
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