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.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given an undirected tree with n nodes, numbered from 0 to n - 1. It is represented by a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.
You are also given two binary strings start and target of length n. For each node x, start[x] is its initial color and target[x] is its desired color.
In one operation, you may pick an edge with index i and toggle both of its endpoints. That is, if the edge is [u, v], then the colors of nodes u and v each flip from '0' to '1' or from '1' to '0'.
Return an array of edge indices whose operations transform start into target. Among all valid sequences with minimum possible length, return the edge indices in increasing order.
If it is impossible to transform start into target, return an array containing a single element equal to -1.
Example 1:
Input: n = 3, edges = [[0,1],[1,2]], start = "010", target = "100"
Output: [0]
Explanation:
Toggle edge with index 0, which flips nodes 0 and 1.
The string changes from "010" to "100", matching the target.
Example 2:
Input: n = 7, edges = [[0,1],[1,2],[2,3],[3,4],[3,5],[1,6]], start = "0011000", target = "0010001"
Output: [1,2,5]
Explanation:
After these operations, the resulting string becomes "0010001", which matches the target.
Example 3:
Input: n = 2, edges = [[0,1]], start = "00", target = "01"
Output: [-1]
Explanation:
There is no sequence of edge toggles that transforms "00" into "01". Therefore, we return [-1].
Constraints:
2 <= n == start.length == target.length <= 105edges.length == n - 1edges[i] = [ai, bi]0 <= ai, bi < nstart[i] is either '0' or '1'.target[i] is either '0' or '1'.edges represents a valid tree.Problem summary: You are given an undirected tree with n nodes, numbered from 0 to n - 1. It is represented by a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. You are also given two binary strings start and target of length n. For each node x, start[x] is its initial color and target[x] is its desired color. In one operation, you may pick an edge with index i and toggle both of its endpoints. That is, if the edge is [u, v], then the colors of nodes u and v each flip from '0' to '1' or from '1' to '0'. Return an array of edge indices whose operations transform start into target. Among all valid sequences with minimum possible length, return the edge indices in increasing order. If it is impossible to transform start into target, return an array containing a single element equal to -1.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree · Topological Sort
3 [[0,1],[1,2]] "010" "100"
7 [[0,1],[1,2],[2,3],[3,4],[3,5],[1,6]] "0011000" "0010001"
2 [[0,1]] "00" "01"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3812: Minimum Edge Toggles on a Tree
class Solution {
private final List<Integer> ans = new ArrayList<>();
private List<int[]>[] g;
private char[] start;
private char[] target;
public List<Integer> minimumFlips(int n, int[][] edges, String start, String target) {
g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
for (int i = 0; i < n - 1; ++i) {
int a = edges[i][0], b = edges[i][1];
g[a].add(new int[] {b, i});
g[b].add(new int[] {a, i});
}
this.start = start.toCharArray();
this.target = target.toCharArray();
if (dfs(0, -1)) {
return List.of(-1);
}
Collections.sort(ans);
return ans;
}
private boolean dfs(int a, int fa) {
boolean rev = start[a] != target[a];
for (var e : g[a]) {
int b = e[0], i = e[1];
if (b != fa && dfs(b, a)) {
ans.add(i);
rev = !rev;
}
}
return rev;
}
}
// Accepted solution for LeetCode #3812: Minimum Edge Toggles on a Tree
func minimumFlips(n int, edges [][]int, start string, target string) []int {
g := make([][]struct{ to, idx int }, n)
for i := 0; i < n-1; i++ {
a, b := edges[i][0], edges[i][1]
g[a] = append(g[a], struct{ to, idx int }{b, i})
g[b] = append(g[b], struct{ to, idx int }{a, i})
}
ans := []int{}
var dfs func(a, fa int) bool
dfs = func(a, fa int) bool {
rev := start[a] != target[a]
for _, p := range g[a] {
b, i := p.to, p.idx
if b != fa && dfs(b, a) {
ans = append(ans, i)
rev = !rev
}
}
return rev
}
if dfs(0, -1) {
return []int{-1}
}
sort.Ints(ans)
return ans
}
# Accepted solution for LeetCode #3812: Minimum Edge Toggles on a Tree
class Solution:
def minimumFlips(
self, n: int, edges: List[List[int]], start: str, target: str
) -> List[int]:
g = [[] for _ in range(n)]
for i, (a, b) in enumerate(edges):
g[a].append((b, i))
g[b].append((a, i))
ans = []
def dfs(a: int, fa: int) -> bool:
rev = start[a] != target[a]
for b, i in g[a]:
if b != fa and dfs(b, a):
ans.append(i)
rev = not rev
return rev
if dfs(0, -1):
return [-1]
ans.sort()
return ans
// Accepted solution for LeetCode #3812: Minimum Edge Toggles on a Tree
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #3812: Minimum Edge Toggles on a Tree
// class Solution {
// private final List<Integer> ans = new ArrayList<>();
// private List<int[]>[] g;
// private char[] start;
// private char[] target;
//
// public List<Integer> minimumFlips(int n, int[][] edges, String start, String target) {
// g = new List[n];
// Arrays.setAll(g, k -> new ArrayList<>());
// for (int i = 0; i < n - 1; ++i) {
// int a = edges[i][0], b = edges[i][1];
// g[a].add(new int[] {b, i});
// g[b].add(new int[] {a, i});
// }
// this.start = start.toCharArray();
// this.target = target.toCharArray();
// if (dfs(0, -1)) {
// return List.of(-1);
// }
// Collections.sort(ans);
// return ans;
// }
//
// private boolean dfs(int a, int fa) {
// boolean rev = start[a] != target[a];
// for (var e : g[a]) {
// int b = e[0], i = e[1];
// if (b != fa && dfs(b, a)) {
// ans.add(i);
// rev = !rev;
// }
// }
// return rev;
// }
// }
// Accepted solution for LeetCode #3812: Minimum Edge Toggles on a Tree
function minimumFlips(n: number, edges: number[][], start: string, target: string): number[] {
const g: number[][][] = Array.from({ length: n }, () => []);
for (let i = 0; i < n - 1; i++) {
const [a, b] = edges[i];
g[a].push([b, i]);
g[b].push([a, i]);
}
const ans: number[] = [];
const dfs = (a: number, fa: number): boolean => {
let rev = start[a] !== target[a];
for (const [b, i] of g[a]) {
if (b !== fa && dfs(b, a)) {
ans.push(i);
rev = !rev;
}
}
return rev;
};
if (dfs(0, -1)) {
return [-1];
}
ans.sort((x, y) => x - y);
return ans;
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.