all-nodes-distance-k-in-binary-tree.py

# Time:  O(n)
# Space: O(n)

# We are given a binary tree (with root node root), a target node,
# and an integer value `K`.
#
# Return a list of the values of all nodes that have a distance K
# from the target node.  The answer can be returned 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.
#
# Note that the inputs "root" and "target" are actually TreeNodes.
# The descriptions of the inputs above are
# just serializations of these objects.
#
# Note:
# - The given tree is non-empty.
# - Each node in the tree has unique values 0 <= node.val <= 500.
# - The target node is a node in the tree.
# - 0 <= K <= 1000.
#
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

import collections

try:
    xrange          # Python 2
except NameError:
    xrange = range  # Python 3


class Solution(object):
    def distanceK(self, root, target, K):
        """
        :type root: TreeNode
        :type target: TreeNode
        :type K: int
        :rtype: List[int]
        """
        def dfs(parent, child, neighbors):
            if not child:
                return
            if parent:
                neighbors[parent.val].append(child.val)
                neighbors[child.val].append(parent.val)
            dfs(child, child.left, neighbors)
            dfs(child, child.right, neighbors)

        neighbors = collections.defaultdict(list)
        dfs(None, root, neighbors)
        bfs = [target.val]
        lookup = set(bfs)
        for _ in xrange(K):
            bfs = [nei for node in bfs
                   for nei in neighbors[node]
                   if nei not in lookup]
            lookup |= set(bfs)
        return bfs