LeetCode #3203 — HARD

Find Minimum Diameter After Merging Two Trees

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

There exist two undirected trees with n and m nodes, numbered from 0 to n - 1 and from 0 to m - 1, respectively. You are given two 2D integer arrays edges1 and edges2 of lengths n - 1 and m - 1, respectively, where edges1[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the first tree and edges2[i] = [u_i, v_i] indicates that there is an edge between nodes u_i and v_i in the second tree.

You must connect one node from the first tree with another node from the second tree with an edge.

Return the minimum possible diameter of the resulting tree.

The diameter of a tree is the length of the longest path between any two nodes in the tree.

Example 1:

Input: edges1 = [[0,1],[0,2],[0,3]], edges2 = [[0,1]]

Output: 3

Explanation:

We can obtain a tree of diameter 3 by connecting node 0 from the first tree with any node from the second tree.

Example 2:

Input: edges1 = [[0,1],[0,2],[0,3],[2,4],[2,5],[3,6],[2,7]], edges2 = [[0,1],[0,2],[0,3],[2,4],[2,5],[3,6],[2,7]]

Output: 5

Explanation:

We can obtain a tree of diameter 5 by connecting node 0 from the first tree with node 0 from the second tree.

Constraints:

1 <= n, m <= 10⁵
edges1.length == n - 1
edges2.length == m - 1
edges1[i].length == edges2[i].length == 2
edges1[i] = [a_i, b_i]
0 <= a_i, b_i < n
edges2[i] = [u_i, v_i]
0 <= u_i, v_i < m
The input is generated such that edges1 and edges2 represent valid trees.

Step 01

Brute Force Baseline

Problem summary: There exist two undirected trees with n and m nodes, numbered from 0 to n - 1 and from 0 to m - 1, respectively. You are given two 2D integer arrays edges1 and edges2 of lengths n - 1 and m - 1, respectively, where edges1[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the first tree and edges2[i] = [ui, vi] indicates that there is an edge between nodes ui and vi in the second tree. You must connect one node from the first tree with another node from the second tree with an edge. Return the minimum possible diameter of the resulting tree. The diameter of a tree is the length of the longest path between any two nodes in the tree.

Baseline thinking

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

Pattern signal: Tree

Example 1

[[0,1],[0,2],[0,3]]
[[0,1]]

Example 2

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

Core Insight

What unlocks the optimal approach

Suppose that we connected node <code>a</code> in tree1 with node <code>b</code> in tree2. The diameter length of the resulting tree will be the largest of the following 3 values: <ol> <li>The diameter of tree 1.</li> <li>The diameter of tree 2.</li> <li>The length of the longest path that starts at node <code>a</code> and that is completely within Tree 1 + The length of the longest path that starts at node <code>b</code> and that is completely within Tree 2 + 1.</li> </ol> The added one in the third value is due to the additional edge that we have added between trees 1 and 2.
Values 1 and 2 are constant regardless of our choice of <code>a</code> and <code>b</code>. Therefore, we need to pick <code>a</code> and <code>b</code> in such a way that minimizes value 3.
If we pick <code>a</code> and <code>b</code> optimally, they will be in the diameters of Tree 1 and Tree 2, respectively. Exactly which nodes of the diameter should we pick?
<code>a</code> is the center of the diameter of tree 1, and <code>b</code> is the center of the diameter of tree 2.

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

Largest constraint values

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 #3203: Find Minimum Diameter After Merging Two Trees
class Solution {
    private List<Integer>[] g;
    private int ans;
    private int a;

    public int minimumDiameterAfterMerge(int[][] edges1, int[][] edges2) {
        int d1 = treeDiameter(edges1);
        int d2 = treeDiameter(edges2);
        return Math.max(Math.max(d1, d2), (d1 + 1) / 2 + (d2 + 1) / 2 + 1);
    }

    public int treeDiameter(int[][] edges) {
        int n = edges.length + 1;
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        ans = 0;
        a = 0;
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        dfs(0, -1, 0);
        dfs(a, -1, 0);
        return ans;
    }

    private void dfs(int i, int fa, int t) {
        for (int j : g[i]) {
            if (j != fa) {
                dfs(j, i, t + 1);
            }
        }
        if (ans < t) {
            ans = t;
            a = i;
        }
    }
}

// Accepted solution for LeetCode #3203: Find Minimum Diameter After Merging Two Trees
func minimumDiameterAfterMerge(edges1 [][]int, edges2 [][]int) int {
	d1 := treeDiameter(edges1)
	d2 := treeDiameter(edges2)
	return max(d1, d2, (d1+1)/2+(d2+1)/2+1)
}

func treeDiameter(edges [][]int) (ans int) {
	n := len(edges) + 1
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	a := 0
	var dfs func(i, fa, t int)
	dfs = func(i, fa, t int) {
		for _, j := range g[i] {
			if j != fa {
				dfs(j, i, t+1)
			}
		}
		if ans < t {
			ans = t
			a = i
		}
	}
	dfs(0, -1, 0)
	dfs(a, -1, 0)
	return
}

# Accepted solution for LeetCode #3203: Find Minimum Diameter After Merging Two Trees
class Solution:
    def minimumDiameterAfterMerge(
        self, edges1: List[List[int]], edges2: List[List[int]]
    ) -> int:
        d1 = self.treeDiameter(edges1)
        d2 = self.treeDiameter(edges2)
        return max(d1, d2, (d1 + 1) // 2 + (d2 + 1) // 2 + 1)

    def treeDiameter(self, edges: List[List[int]]) -> int:
        def dfs(i: int, fa: int, t: int):
            for j in g[i]:
                if j != fa:
                    dfs(j, i, t + 1)
            nonlocal ans, a
            if ans < t:
                ans = t
                a = i

        g = defaultdict(list)
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        ans = a = 0
        dfs(0, -1, 0)
        dfs(a, -1, 0)
        return ans

// Accepted solution for LeetCode #3203: Find Minimum Diameter After Merging Two Trees
impl Solution {
    pub fn minimum_diameter_after_merge(edges1: Vec<Vec<i32>>, edges2: Vec<Vec<i32>>) -> i32 {
        let d1 = Self::tree_diameter(&edges1);
        let d2 = Self::tree_diameter(&edges2);
        d1.max(d2).max((d1 + 1) / 2 + (d2 + 1) / 2 + 1)
    }

    fn tree_diameter(edges: &Vec<Vec<i32>>) -> i32 {
        let n = edges.len() + 1;
        let mut g = vec![vec![]; n];
        for e in edges {
            let a = e[0] as usize;
            let b = e[1] as usize;
            g[a].push(b);
            g[b].push(a);
        }
        let mut ans = 0;
        let mut a = 0;
        fn dfs(g: &Vec<Vec<usize>>, i: usize, fa: isize, t: i32, ans: &mut i32, a: &mut usize) {
            for &j in &g[i] {
                if j as isize != fa {
                    dfs(g, j, i as isize, t + 1, ans, a);
                }
            }
            if *ans < t {
                *ans = t;
                *a = i;
            }
        }
        dfs(&g, 0, -1, 0, &mut ans, &mut a);
        dfs(&g, a, -1, 0, &mut ans, &mut a);
        ans
    }
}

// Accepted solution for LeetCode #3203: Find Minimum Diameter After Merging Two Trees
function minimumDiameterAfterMerge(edges1: number[][], edges2: number[][]): number {
    const d1 = treeDiameter(edges1);
    const d2 = treeDiameter(edges2);
    return Math.max(d1, d2, Math.ceil(d1 / 2) + Math.ceil(d2 / 2) + 1);
}

function treeDiameter(edges: number[][]): number {
    const n = edges.length + 1;
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    let [ans, a] = [0, 0];
    const dfs = (i: number, fa: number, t: number): void => {
        for (const j of g[i]) {
            if (j !== fa) {
                dfs(j, i, t + 1);
            }
        }
        if (ans < t) {
            ans = t;
            a = i;
        }
    };
    dfs(0, -1, 0);
    dfs(a, -1, 0);
    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 + m)

Space

O(n + m)

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.

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.