Leetcode #863: All Nodes Distance K in Binary Tree
In this guide, we solve Leetcode #863 All Nodes Distance K in Binary Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Hash Table, Binary Tree
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
# 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]:
def dfs(root, fa):
if root is None:
return
g[root] = fa
dfs(root.left, root)
dfs(root.right, root)
def dfs2(root, fa, k):
if root is None:
return
if k == 0:
ans.append(root.val)
return
for nxt in (root.left, root.right, g[root]):
if nxt != fa:
dfs2(nxt, root, k - 1)
g = {}
dfs(root, None)
ans = []
dfs2(target, None, k)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.