<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 {
    private:
    bool hasCycle(int node, vector<int>& lC, vector<int>& rC, vector<bool>& vis) {

        if (node == -1) return false;
        if (vis[node]) return true;
        vis[node] = true;

        bool leftCycle = hasCycle(lC[node], lC, rC, vis);
        bool rightCycle = hasCycle(rC[node], lC, rC, vis);

        return leftCycle || rightCycle;
    }
public:
    bool validateBinaryTreeNodes(int n, vector<int>& lC, vector<int>& rC) {
        
        unordered_map<int,int> mp;
        for (int node=0 ; node<n ; node++) {
            if (lC[node] != -1) {
                if (mp.find(lC[node]) != mp.end()) return false;
                mp[lC[node]] = node;
            }

            if (rC[node] != -1) {
                if (mp.find(rC[node]) != mp.end()) return false;
                mp[rC[node]] = node;
            }
        }

        int root = -1;
        for (int node=0 ; node<n ; node++) {
            if (mp.find(node) == mp.end()) {
                if (root != -1) return false;
                root = node;
            }
        }

        if (root == -1) return false;

        vector<bool> vis(n, false);
        if (hasCycle(root, lC, rC, vis)) return false;

        for (int node=0 ; node<n ; node++) {
            if (!vis[node]) return false;
        }

        return true;
    }
};
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