863-all-nodes-distance-k-in-binary-tree

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Description

Given the root of a binary tree, the value of a target node target, and an integer k, return an array of the values of all nodes that have a distance k from the target node.

You can return the answer in any order.

 

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, k = 2
Output: [7,4,1]
Explanation: The nodes that are a distance 2 from the target node (with value 5) have values 7, 4, and 1.

Example 2:

Input: root = [1], target = 1, k = 3
Output: []

 

Constraints:

• The number of nodes in the tree is in the range [1, 500].
• 0 <= Node.val <= 500
• All the values Node.val are unique.
• target is the value of one of the nodes in the tree.
• 0 <= k <= 1000

Solution(Python)

class Solution(object):
    def distanceK(self, root, target, K):
        def dfs(node, par=None):
            if node:
                node.par = par
                dfs(node.left, node)
                dfs(node.right, node)

        dfs(root)

        queue = collections.deque([(target, 0)])
        seen = {target}
        while queue:
            if queue[0][1] == K:
                return [node.val for node, d in queue]
            node, d = queue.popleft()
            for nei in (node.left, node.right, node.par):
                if nei and nei not in seen:
                    seen.add(nei)
                    queue.append((nei, d + 1))

        return []