<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:
 void dfs(int node, vector<int> adj[], vector<int>& vis) {
        vis[node] = 1;
        for (auto it : adj[node]) {
            if (!vis[it]) {
                dfs(it, adj, vis);
            }
        }
    }
 // Cycle detection in a directed graph
    bool iscycle(int node, vector<int> adj[], vector<int>& vis, vector<int>& pathVis) {
        vis[node] = 1;
        pathVis[node] = 1;
        for (auto it : adj[node]) {
            if (!vis[it]) {
                if (iscycle(it, adj, vis, pathVis)) return true;
            } else if (pathVis[it]) {
                return true;  // cycle detected
            }
        }
        pathVis[node] = 0;  // backtrack
        return false;
    }
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        vector<int> indeg(n, 0);  // track indegree of nodes
        vector<int> adj[n];       // adjacency list

        // Build the adjacency list and indegree array (directed graph)
        for (int i = 0; i < n; i++) {
            if (leftChild[i] != -1) {
                indeg[leftChild[i]]++;
                adj[i].push_back(leftChild[i]);  // directed edge: i -> leftChild[i]
            }
            if (rightChild[i] != -1) {
                indeg[rightChild[i]]++;
                adj[i].push_back(rightChild[i]);  // directed edge: i -> rightChild[i]
            }
        }

        // Count nodes with indegree 0 (root candidates)
        int rootCount = 0;
        int rootNode = -1;
        for (int i = 0; i < n; i++) {
            if (indeg[i] == 0) {
                rootCount++;
                rootNode = i;
            }
            if (indeg[i] > 1) return false;  // A node with more than one parent is invalid
        }

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

        // Check for cycles using DFS
       // vector<int> vis(n, 0), pathVis(n, 0);
        //if (iscycle(rootNode, adj, vis, pathVis)) return false;

        // Ensure all nodes are visited (single connected component)
        vector<int> visDfs(n, 0);
        dfs(rootNode, adj, visDfs);
        for (int i = 0; i < n; i++) {
            if (!visDfs[i]) return false;
        }

        return true;
    }
};
43692