<p>You have <code>n</code> binary tree nodes numbered from <code>0</code> to <code>n - 1</code> where node <code>i</code> has two children <code>leftChild[i]</code> and <code>rightChild[i]</code>, return <code>true</code> if and only if <strong>all</strong> the given nodes form <strong>exactly one</strong> valid binary tree.</p>

<p>If node <code>i</code> has no left child then <code>leftChild[i]</code> will equal <code>-1</code>, similarly for the right child.</p>

<p>Note that the nodes have no values and that we only use the node numbers in this problem.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/08/23/1503_ex1.png" style="width: 195px; height: 287px;" />
<pre>
<strong>Input:</strong> n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/08/23/1503_ex2.png" style="width: 183px; height: 272px;" />
<pre>
<strong>Input:</strong> n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/08/23/1503_ex3.png" style="width: 82px; height: 174px;" />
<pre>
<strong>Input:</strong> n = 2, leftChild = [1,0], rightChild = [-1,-1]
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>n == leftChild.length == rightChild.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>-1 &lt;= leftChild[i], rightChild[i] &lt;= n - 1</code></li>
</ul>

class Solution {
public:
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
    vector<int> inDegree(n, 0);

    // Calculate in-degrees to identify the root and check for multiple parents
    for (int i = 0; i < n; ++i) {
        if (leftChild[i] != -1) {
            inDegree[leftChild[i]]++;
        }
        if (rightChild[i] != -1) {
            inDegree[rightChild[i]]++;
        }
    }

    int rootCount = 0, root = -1;

    // Find the root and check for multiple roots
    for (int i = 0; i < n; ++i) {
        if (inDegree[i] == 0) {
            root = i;
            rootCount++;
        }
        else if (inDegree[i] > 1) {
            return false; // A node has more than one parent
        }
    }

    if (rootCount != 1) {
        return false; // There should be exactly one root
    }

    unordered_set<int> visited;

    // Run DFS from the root and check for cycles
    if (!dfs(root, visited, leftChild, rightChild)) {
        return false;
    }

    // Check if all nodes are visited (i.e., the graph is connected)
    return visited.size() == n;
}

bool dfs(int curr, unordered_set<int>& visited, vector<int>& leftChild, vector<int>& rightChild) {
    if (curr == -1) {
        return true;
    }

    if (visited.count(curr)) {
        return false; // Found a cycle
    }

    visited.insert(curr);

    // Visit left and right children
    return dfs(leftChild[curr], visited, leftChild, rightChild) &&
        dfs(rightChild[curr], visited, leftChild, rightChild);
}

};
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