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