leetcode-863-All-Nodes-Distance-K-in-Binary-Tree

TOREVIEW

描述


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:

img

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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:

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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

分析


先根据树建一个图,然后用深度优先遍历,获取深度。

解决方案1(Python)


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# 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]:
nodeParentDict = dict()

def dfsFindParent(node):
if node:
if node.left:
nodeParentDict[node.left] = node
if node.right:
nodeParentDict[node.right] = node
dfsFindParent(node.left)
dfsFindParent(node.right)

dfsFindParent(root)

if k == 0:
return [target.val]

result = []

queue = collections.deque()
visited = set()
queue.append(target)
visited.add(target)
level = 0

while queue and level < k:
level += 1
for _ in range(len(queue)):
now = queue.popleft()
for nextNode in [nodeParentDict[now] if now in nodeParentDict else None, now.left, now.right]:
if nextNode and nextNode not in visited:
if level == k:
result.append(nextNode.val)
queue.append(nextNode)
visited.add(nextNode)
return result

题目来源