<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:

    int traverse(int idx, vector<int>& vis, vector<vector<int>>& adj, int par) {

        if(vis[idx])
            return 0;
        
        vis[idx] = 1;
        int ans = 1;

        for(auto it: adj[idx]) {
            
            if(it == par)
                continue;

            ans &= traverse(it, vis, adj, idx);
        }

        return ans;
    }

    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        
        vector<int> vis(n);
        vector<vector<int>> adj(n);

        int edges = 0;
        vector<int> indeg(n);
        for(int i=0;i<n;i++) {
            if(leftChild[i] != -1) {
                edges++;
                indeg[leftChild[i]]++;
                adj[i].push_back(leftChild[i]);
                adj[leftChild[i]].push_back(i);
            }
            if(rightChild[i] != -1) {
                edges++;
                indeg[rightChild[i]]++;
                adj[i].push_back(rightChild[i]);
                adj[rightChild[i]].push_back(i);
            }
        }

        if(edges != n-1)
            return 0;

        
        int ans = traverse(0, vis, adj, -1);

        for(int i=0;i<n;i++) {
            ans &= vis[i];

            if(indeg[i] > 1)
                ans = 0;

        }

        return ans;
    }
};
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