<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:
    unordered_map<int, int> parent;

    bool findParent(int n, vector<int>& leftChild, vector<int>& rightChild) {
        #pragma unroll
        for (int i=0; i < n; i++) {
            int left = leftChild[i], right=rightChild[i];
            if (left != -1) {
                if (parent.count(left) != 0 || left == i) {
                    return 0;
                }
                parent[left] = i;
            }
            if (right != -1) {
                if (parent.count(right) != 0 || right == i) {
                    return 0;
                }
                parent[right] = i;
            }
        }
        return 1;
    }

    vector<int> root;

    int findRoot(int i) {
        if (parent.count(i)==0) 
            return i;
        
        if (root[i]!=-1) return root[i];
        // Mark the node as visited during this call
        root[i]=-2; //for detecting cycle
        return root[i]=findRoot(parent[i]);
    }

    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        bool yes = findParent(n, leftChild, rightChild);
        if (!yes) return 0;
        root.assign(n, -1);
        int R0=findRoot(0);
        if (R0==-2||parent.count(R0)) return 0;
        #pragma unroll
        for (int i = 0; i<n; i++) {
            if (root[i]==-1) {
                int R=findRoot(i);
                if (R==-2||R!=R0||parent.count(R)) return 0;//Root shouldn't have a parent.
            }
        }
        return 1;
    }
};
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