<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 bfs(map<int, vector<int>>& adj, int root, int n) {
        queue<int> q;
        vector<bool> visited(n, false);
        q.push(root);
        visited[root] = true;
        int count = 0;
        
        while (!q.empty()) {
            int u = q.front();
            q.pop();
            count++;
            
            for (auto v : adj[u]) {
                if (visited[v]) return false; // Detects a cycle or multiple parents for a node
                visited[v] = true;
                q.push(v);
            }
        }
        
        return count == n; // Ensure all nodes are visited
    }
    
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        map<int, vector<int>> adj;
        vector<int> parent(n, -1);
        
        for (int i = 0; i < n; i++) {
            int l = leftChild[i];
            int r = rightChild[i];
            
            if (l != -1) {
                if (parent[l] != -1) return false; // A child already has a parent
                parent[l] = i;
                adj[i].push_back(l);
            }
            
            if (r != -1) {
                if (parent[r] != -1) return false; // A child already has a parent
                parent[r] = i;
                adj[i].push_back(r);
            }
        }
        
        int root = -1;
        for (int i = 0; i < n; i++) {
            if (parent[i] == -1) {
                if (root != -1) return false; // More than one root found
                root = i;
            }
        }
        
        if (root == -1) return false; // No root found
        
        return bfs(adj, root, n);
    }
};

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