<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>

class Solution:
    def distanceK(self, root: TreeNode, target: TreeNode, k: int) -> List[int]:
        graph = collections.defaultdict(list)
        
        # Recursively build the undirected graph from the given binary tree.
        def build_graph(cur, parent):
            if cur and parent:
                graph[cur.val].append(parent.val)
                graph[parent.val].append(cur.val)
            if cur.left:
                build_graph(cur.left, cur)
            if cur.right:
                build_graph(cur.right, cur) 
        build_graph(root, None)
        

        answer = []
        visited = set([target.val])
        
        def dfs(cur, distance):
            if distance == k:
                answer.append(cur)
                return
            for neighbor in graph[cur]:
                if neighbor not in visited:
                    visited.add(neighbor)
                    dfs(neighbor, distance + 1)
        dfs(target.val, 0)
        
        return answer
16700