<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>

#include <vector>
#include <deque>
#include <unordered_set>

class Solution {
public:
    bool validateBinaryTreeNodes(int n, std::vector<int>& leftChild, std::vector<int>& rightChild) {
        std::deque<int> queue; // Create a queue for BFS traversal.
        std::unordered_set<int> visited; // Create a set to track visited nodes.
        int root = findRoot(n, leftChild, rightChild); // Find the root of the tree.
        if (root == -1) return false; // If there is no valid root, return false.

        queue.push_back(root); // Start the BFS from the root node.
        while (!queue.empty()) {
            int node = queue.front(); // Dequeue a node for processing.
            queue.pop_front();
            if (node == -1) continue; // Skip null nodes.

            if (!visited.insert(node).second) return false; // If the node is visited again, return false.

            int left = leftChild[node];
            int right = rightChild[node];

            queue.push_back(left); // Enqueue the left child.
            queue.push_back(right); // Enqueue the right child.
        }

        return visited.size() == n; // Check if all nodes have been visited once (i.e., one connected component).
    }

private:
    int findRoot(int n, std::vector<int>& left, std::vector<int>& right) {
        std::unordered_set<int> set;
        for (int i = 0; i < n; i++) set.insert(i); // Initialize a set with all node indices.
        for (int node : left) set.erase(node); // Remove left child nodes from the set.
        for (int node : right) set.erase(node); // Remove right child nodes from the set.

        return set.size() == 1 ? *set.begin() : -1; // Return the root node if there's only one, or -1 if none or multiple.
    }
};
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