<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]:
        ans = []
        parent = {}
        queue = deque()
        queue.append(root)
        while queue:
            size = len(queue)
            for _ in range(size):
                top = queue.popleft()
                if top.left:
                    parent[top.left.val] = top
                    queue.append(top.left)
                if top.right:
                    parent[top.right.val] = top
                    queue.append(top.right)
        visited = {}
        queue.append(target)
        while k > 0 and queue:
            size = len(queue)
            for _ in range(size):
                top = queue.popleft()
                visited[top.val] = 1
                if top.left and top.left.val not in visited:
                    queue.append(top.left)

                if top.right and top.right.val not in visited:
                    queue.append(top.right)

                if top.val in parent and parent[top.val].val not in visited:
                    queue.append(parent[top.val])
            k -= 1
        while queue:
            ans.append(queue.popleft().val)

        return ans
        
16600