LeetCode #863 — MEDIUM

All Nodes Distance K in Binary Tree

Move from brute-force thinking to an efficient approach using hash map strategy.

The Problem

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.

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.

Example 2:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Tree

Example 1

[3,5,1,6,2,0,8,null,null,7,4]
5
2

Example 2

[1]
1
3

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #863: All Nodes Distance K in Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private Map<TreeNode, TreeNode> g = new HashMap<>();
    private List<Integer> ans = new ArrayList<>();

    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        dfs(root, null);
        dfs2(target, null, k);
        return ans;
    }

    private void dfs(TreeNode root, TreeNode fa) {
        if (root == null) {
            return;
        }
        g.put(root, fa);
        dfs(root.left, root);
        dfs(root.right, root);
    }

    private void dfs2(TreeNode root, TreeNode fa, int k) {
        if (root == null) {
            return;
        }
        if (k == 0) {
            ans.add(root.val);
            return;
        }
        for (TreeNode nxt : new TreeNode[] {root.left, root.right, g.get(root)}) {
            if (nxt != fa) {
                dfs2(nxt, root, k - 1);
            }
        }
    }
}

// Accepted solution for LeetCode #863: All Nodes Distance K in Binary Tree
func distanceK(root *TreeNode, target *TreeNode, k int) []int {
	g := make(map[*TreeNode]*TreeNode)
	ans := []int{}

	var dfs func(node, fa *TreeNode)
	dfs = func(node, fa *TreeNode) {
		if node == nil {
			return
		}
		g[node] = fa
		dfs(node.Left, node)
		dfs(node.Right, node)
	}

	var dfs2 func(node, fa *TreeNode, k int)
	dfs2 = func(node, fa *TreeNode, k int) {
		if node == nil {
			return
		}
		if k == 0 {
			ans = append(ans, node.Val)
			return
		}
		for _, nxt := range []*TreeNode{node.Left, node.Right, g[node]} {
			if nxt != fa {
				dfs2(nxt, node, k-1)
			}
		}
	}

	dfs(root, nil)
	dfs2(target, nil, k)

	return ans
}

# Accepted solution for LeetCode #863: All Nodes Distance K in Binary Tree
# 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

// Accepted solution for LeetCode #863: All Nodes Distance K in Binary Tree
struct Solution;
use rustgym_util::*;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;

impl Solution {
    fn distance_k(root: TreeLink, p: TreeLink, k: i32) -> Vec<i32> {
        let k = k as usize;
        let mut adj: HashMap<i32, Vec<i32>> = HashMap::new();
        Self::dfs(root, &mut adj);
        let mut res = vec![];
        let mut queue: VecDeque<(i32, usize)> = VecDeque::new();
        let start = p.unwrap().borrow().val;
        let mut visited: HashSet<i32> = HashSet::new();
        visited.insert(start);
        queue.push_back((start, 0));
        while let Some((u, d)) = queue.pop_front() {
            if d == k {
                res.push(u);
            } else {
                for &mut v in adj.entry(u).or_default() {
                    if visited.insert(v) {
                        queue.push_back((v, d + 1));
                    }
                }
            }
        }
        res
    }

    fn dfs(root: TreeLink, adj: &mut HashMap<i32, Vec<i32>>) {
        if let Some(node) = root {
            let mut node = node.borrow_mut();
            let u = node.val;
            let left = node.left.take();
            let right = node.right.take();
            if left.is_some() {
                let v = left.as_ref().unwrap().borrow().val;
                adj.entry(u).or_default().push(v);
                adj.entry(v).or_default().push(u);
                Self::dfs(left, adj);
            }
            if right.is_some() {
                let v = right.as_ref().unwrap().borrow().val;
                adj.entry(u).or_default().push(v);
                adj.entry(v).or_default().push(u);
                Self::dfs(right, adj);
            }
        }
    }
}

#[test]
fn test() {
    let root = tree!(
        3,
        tree!(5, tree!(6), tree!(2, tree!(7), tree!(4))),
        tree!(1, tree!(0), tree!(8))
    );
    let p = tree!(5);
    let k = 2;
    let mut res = vec![7, 4, 1];
    let mut ans = Solution::distance_k(root, p, k);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #863: All Nodes Distance K in Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function distanceK(root: TreeNode | null, target: TreeNode | null, k: number): number[] {
    const g = new Map<TreeNode, TreeNode | null>();
    const ans: number[] = [];

    const dfs = (node: TreeNode | null, fa: TreeNode | null) => {
        if (!node) {
            return;
        }
        g.set(node, fa);
        dfs(node.left, node);
        dfs(node.right, node);
    };

    const dfs2 = (node: TreeNode | null, fa: TreeNode | null, k: number) => {
        if (!node) {
            return;
        }
        if (k === 0) {
            ans.push(node.val);
            return;
        }
        for (const nxt of [node.left, node.right, g.get(node) || null]) {
            if (nxt !== fa) {
                dfs2(nxt, node, k - 1);
            }
        }
    };

    dfs(root, null);
    dfs2(target, null, k);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.