<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 n;
    int parent[10004];

    int findParent(int node) {
        if(node == parent[node]) return node;
        return parent[node] = findParent(parent[node]);
    }

    void Union(int u , int v) {
        int parent_u = findParent(u);
        int parent_v = findParent(v);

        parent[parent_v] = parent_u;
    }

    bool validateBinaryTreeNodes(int nn, vector<int>& leftChild, vector<int>& rightChild) {
        n = nn;
        int components = n;

        for(int i=0;i<n;i++) parent[i] = i;

        for(int i=0;i<n;i++) {
            int node = i;
            int leftC = leftChild[i];
            int rightC = rightChild[i];

            if(leftC != -1) {
                if(findParent(node) == findParent(leftC)) return false;
                if(findParent(leftC) != leftC) return false;
                Union(node, leftC);
                components--;
            }

            if(rightC != -1) {
                if(findParent(node) == findParent(rightC)) return false;
                if(findParent(rightC) != rightC) return false;
                Union(node, rightC);
                components--;
            }
        }
        
        if(components > 1) return false;

        return true;
    }
};
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