Problem description
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.
Solution
- We can first use DFS to add a parent attr to all the tree nodes.
- Then we can use a second DFS to find all the nodes that have K distance from target.
Below is my python implementation
# 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, target, K):
"""
:type root: TreeNode
:type target: TreeNode
:type K: int
:rtype: List[int]
"""
ans = []
def dfs1(root, parent)->None:#add parent attr to the tree node
if root == None:
return
root.parent = parent
dfs1(root.left, root)
dfs1(root.right, root)
return
visited = {}
def dfs2(node, K)->None:
if not node:
return
visited[node] = True
if K == 0:
ans.append(node.val)
if node.left not in visited:
dfs2(node.left, K - 1)
if node.right not in visited:
dfs2(node.right, K - 1)
if node.parent not in visited:
dfs2(node.parent, K - 1)
return
dfs1(root, None)
dfs2(target, K)
return ans