<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.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def distanceK(self, root: TreeNode, target: TreeNode, k: int) -> List[int]:
        if not target:
            return []
        parents = {}
        def build_parents(cur, p):
            if not cur:
                return
            parents[cur] = p
            build_parents(cur.left, cur)
            build_parents(cur.right, cur)
        build_parents(root, None)
        queue = deque([target])
        traverse = 0
        result = []
        seen = set()
        while queue:
            if traverse > k:
                break
            for _ in range(len(queue)):
                cur = queue.pop()
                if cur in seen:
                    continue
                if traverse == k:
                    result.append(cur.val)
                    continue
                seen.add(cur)
                if cur.left:
                    queue.appendleft(cur.left)
                if cur.right:
                    queue.appendleft(cur.right)
                if parents.get(cur):
                    queue.appendleft(parents[cur])
            traverse += 1
        return result
    
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