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
.
brute force,求出root到所有node的path,然后求距离
# 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, t, K):
"""
:type root: TreeNode
:type target: TreeNode
:type K: int
:rtype: List[TreeNode]
"""
p, d = [], {}
def dfs(r):
if not r: return
p.append(r)
d[r] = list(p)
dfs(r.left)
dfs(r.right)
p.pop()
dfs(root)
res=[]
for k in d:
i=0
while i<len(d[k]) and i<len(d[t]) and d[k][i]==d[t][i]:
i+=1
if len(d[k])-i + len(d[t])-i == K:
res.append(k.val)
return res