<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 dfs(vector<vector<int>>& graph, int src, vector<bool>& vis, vector<bool>& inPath) {
        vis[src] = true;
        inPath[src] = true; // Mark the node as being in the current path
        
        for (auto nbr : graph[src]) {
            if (inPath[nbr]) return false;  // If we revisit a node in the current path, we have a cycle
            if (!vis[nbr]) {
                if (!dfs(graph, nbr, vis, inPath)) {
                    return false; // If dfs on the neighbor returns false, we propagate the failure
                }
            }
        }
        
        inPath[src] = false;  // Mark the node as out of the current path
        return true;
    }
    
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        vector<vector<int>> graph(n);
        vector<int> indegree(n, 0); // To track the indegree of each node
        
        for (int i = 0; i < n; i++) {
            if (leftChild[i] != -1) {
                graph[i].push_back(leftChild[i]);
                indegree[leftChild[i]]++;
            }
            if (rightChild[i] != -1) {
                graph[i].push_back(rightChild[i]);
                indegree[rightChild[i]]++;
            }
        }
        
        // Check if there is exactly one root node (indegree 0)
        int rootCount = 0, root = -1;
        for (int i = 0; i < n; i++) {
            if (indegree[i] == 0) {
                rootCount++;
                root = i;
            } else if (indegree[i] > 1) {
                return false; // Invalid tree, a node cannot have more than one parent
            }
        }
        
        if (rootCount != 1) return false; // There must be exactly one root
        
        vector<bool> vis(n, false), inPath(n, false);
        if (!dfs(graph, root, vis, inPath)) return false; // Run DFS from the root
        
        // Check if all nodes are visited (the tree must be fully connected)
        for (bool v : vis) {
            if (!v) return false;
        }
        
        return true;
    }
};

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