Maximize the Number of Target Nodes After Connecting Trees II
<p>There exist two <strong>undirected </strong>trees with <code>n</code> and <code>m</code> nodes, labeled from <code>[0, n - 1]</code> and <code>[0, m - 1]</code>, respectively.</p>

<p>You are given two 2D integer arrays <code>edges1</code> and <code>edges2</code> of lengths <code>n - 1</code> and <code>m - 1</code>, respectively, where <code>edges1[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the first tree and <code>edges2[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> in the second tree.</p>

<p>Node <code>u</code> is <strong>target</strong> to node <code>v</code> if the number of edges on the path from <code>u</code> to <code>v</code> is even.&nbsp;<strong>Note</strong> that a node is <em>always</em> <strong>target</strong> to itself.</p>

<p>Return an array of <code>n</code> integers <code>answer</code>, where <code>answer[i]</code> is the <strong>maximum</strong> possible number of nodes that are <strong>target</strong> to node <code>i</code> of the first tree if you had to connect one node from the first tree to another node in the second tree.</p>

<p><strong>Note</strong> that queries are independent from each other. That is, for every query you will remove the added edge before proceeding to the next query.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">edges1 = [[0,1],[0,2],[2,3],[2,4]], edges2 = [[0,1],[0,2],[0,3],[2,7],[1,4],[4,5],[4,6]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[8,7,7,8,8]</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>For <code>i = 0</code>, connect node 0 from the first tree to node 0 from the second tree.</li>
	<li>For <code>i = 1</code>, connect node 1 from the first tree to node 4 from the second tree.</li>
	<li>For <code>i = 2</code>, connect node 2 from the first tree to node 7 from the second tree.</li>
	<li>For <code>i = 3</code>, connect node 3 from the first tree to node 0 from the second tree.</li>
	<li>For <code>i = 4</code>, connect node 4 from the first tree to node 4 from the second tree.</li>
</ul>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3300-3399/3373.Maximize%20the%20Number%20of%20Target%20Nodes%20After%20Connecting%20Trees%20II/images/3982-1.png" style="width: 600px; height: 169px;" /></div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">edges1 = [[0,1],[0,2],[0,3],[0,4]], edges2 = [[0,1],[1,2],[2,3]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[3,6,6,6,6]</span></p>

<p><strong>Explanation:</strong></p>

<p>For every <code>i</code>, connect node <code>i</code> of the first tree with any node of the second tree.</p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3300-3399/3373.Maximize%20the%20Number%20of%20Target%20Nodes%20After%20Connecting%20Trees%20II/images/3928-2.png" style="height: 281px; width: 500px;" /></div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n, m &lt;= 10<sup>5</sup></code></li>
	<li><code>edges1.length == n - 1</code></li>
	<li><code>edges2.length == m - 1</code></li>
	<li><code>edges1[i].length == edges2[i].length == 2</code></li>
	<li><code>edges1[i] = [a<sub>i</sub>, b<sub>i</sub>]</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>edges2[i] = [u<sub>i</sub>, v<sub>i</sub>]</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt; m</code></li>
	<li>The input is generated such that <code>edges1</code> and <code>edges2</code> represent valid trees.</li>
</ul>
Tree; Depth-First Search; Breadth-First Search
public class Solution {
    public int[] MaxTargetNodes(int[][] edges1, int[][] edges2) {
        var g1 = Build(edges1);
        var g2 = Build(edges2);
        int n = g1.Length, m = g2.Length;
        var c1 = new int[n];
        var c2 = new int[m];
        var cnt1 = new int[2];
        var cnt2 = new int[2];

        Dfs(g2, 0, -1, c2, 0, cnt2);
        Dfs(g1, 0, -1, c1, 0, cnt1);

        int t = Math.Max(cnt2[0], cnt2[1]);
        var ans = new int[n];
        for (int i = 0; i < n; i++) {
            ans[i] = t + cnt1[c1[i]];
        }
        return ans;
    }

    private List<int>[] Build(int[][] edges) {
        int n = edges.Length + 1;
        var g = new List<int>[n];
        for (int i = 0; i < n; i++) {
            g[i] = new List<int>();
        }
        foreach (var e in edges) {
            int a = e[0], b = e[1];
            g[a].Add(b);
            g[b].Add(a);
        }
        return g;
    }

    private void Dfs(List<int>[] g, int a, int fa, int[] c, int d, int[] cnt) {
        c[a] = d;
        cnt[d]++;
        foreach (var b in g[a]) {
            if (b != fa) {
                Dfs(g, b, a, c, d ^ 1, cnt);
            }
        }
    }
}
