Smallest Missing Genetic Value in Each Subtree
<p>There is a <strong>family tree</strong> rooted at <code>0</code> consisting of <code>n</code> nodes numbered <code>0</code> to <code>n - 1</code>. You are given a <strong>0-indexed</strong> integer array <code>parents</code>, where <code>parents[i]</code> is the parent for node <code>i</code>. Since node <code>0</code> is the <strong>root</strong>, <code>parents[0] == -1</code>.</p>

<p>There are <code>10<sup>5</sup></code> genetic values, each represented by an integer in the <strong>inclusive</strong> range <code>[1, 10<sup>5</sup>]</code>. You are given a <strong>0-indexed</strong> integer array <code>nums</code>, where <code>nums[i]</code> is a <strong>distinct </strong>genetic value for node <code>i</code>.</p>

<p>Return <em>an array </em><code>ans</code><em> of length </em><code>n</code><em> where </em><code>ans[i]</code><em> is</em> <em>the <strong>smallest</strong> genetic value that is <strong>missing</strong> from the subtree rooted at node</em> <code>i</code>.</p>

<p>The <strong>subtree</strong> rooted at a node <code>x</code> contains node <code>x</code> and all of its <strong>descendant</strong> nodes.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2003.Smallest%20Missing%20Genetic%20Value%20in%20Each%20Subtree/images/case-1.png" style="width: 204px; height: 167px;" />
<pre>
<strong>Input:</strong> parents = [-1,0,0,2], nums = [1,2,3,4]
<strong>Output:</strong> [5,1,1,1]
<strong>Explanation:</strong> The answer for each subtree is calculated as follows:
- 0: The subtree contains nodes [0,1,2,3] with values [1,2,3,4]. 5 is the smallest missing value.
- 1: The subtree contains only node 1 with value 2. 1 is the smallest missing value.
- 2: The subtree contains nodes [2,3] with values [3,4]. 1 is the smallest missing value.
- 3: The subtree contains only node 3 with value 4. 1 is the smallest missing value.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/2000-2099/2003.Smallest%20Missing%20Genetic%20Value%20in%20Each%20Subtree/images/case-2.png" style="width: 247px; height: 168px;" />
<pre>
<strong>Input:</strong> parents = [-1,0,1,0,3,3], nums = [5,4,6,2,1,3]
<strong>Output:</strong> [7,1,1,4,2,1]
<strong>Explanation:</strong> The answer for each subtree is calculated as follows:
- 0: The subtree contains nodes [0,1,2,3,4,5] with values [5,4,6,2,1,3]. 7 is the smallest missing value.
- 1: The subtree contains nodes [1,2] with values [4,6]. 1 is the smallest missing value.
- 2: The subtree contains only node 2 with value 6. 1 is the smallest missing value.
- 3: The subtree contains nodes [3,4,5] with values [2,1,3]. 4 is the smallest missing value.
- 4: The subtree contains only node 4 with value 1. 2 is the smallest missing value.
- 5: The subtree contains only node 5 with value 3. 1 is the smallest missing value.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> parents = [-1,2,3,0,2,4,1], nums = [2,3,4,5,6,7,8]
<strong>Output:</strong> [1,1,1,1,1,1,1]
<strong>Explanation:</strong> The value 1 is missing from all the subtrees.
</pre>

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

<ul>
	<li><code>n == parents.length == nums.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= parents[i] &lt;= n - 1</code> for <code>i != 0</code></li>
	<li><code>parents[0] == -1</code></li>
	<li><code>parents</code> represents a valid tree.</li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li>Each <code>nums[i]</code> is distinct.</li>
</ul>
Tree; Depth-First Search; Union Find; Dynamic Programming
func smallestMissingValueSubtree(parents []int, nums []int) []int {
	n := len(nums)
	g := make([][]int, n)
	vis := make([]bool, n)
	has := make([]bool, n+2)
	idx := -1
	ans := make([]int, n)
	for i, p := range parents {
		if i > 0 {
			g[p] = append(g[p], i)
		}
		if nums[i] == 1 {
			idx = i
		}
		ans[i] = 1
	}
	if idx < 0 {
		return ans
	}
	var dfs func(int)
	dfs = func(i int) {
		if vis[i] {
			return
		}
		vis[i] = true
		if nums[i] < len(has) {
			has[nums[i]] = true
		}
		for _, j := range g[i] {
			dfs(j)
		}
	}
	for i := 2; idx != -1; idx = parents[idx] {
		dfs(idx)
		for has[i] {
			i++
		}
		ans[idx] = i
	}
	return ans
}