<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:
    def validateBinaryTreeNodes(self, n: int, leftChild: List[int], rightChild: List[int]) -> bool:

        djs = disjointSet(n)

        for i in range(len(leftChild)):
            if leftChild[i]!= -1:
                if djs.root[leftChild[i]] != leftChild[i]:
                    return False
                if djs.connected(i, leftChild[i]):
                    return False
                djs.union(i, leftChild[i])
            
            if rightChild[i]!= -1:
                if djs.root[rightChild[i]] != rightChild[i]:
                    return False
                if djs.connected(i, rightChild[i]):
                    return False
                djs.union(i, rightChild[i])

        z = 0
        for i in range(n):
            x = djs.find(i)
            if x == i:
                z +=1
        
        print(djs.root)

        return z == 1

class disjointSet:

    def __init__(self, n):
        self.n = 0
        self.root = [i for i in range(n)]
        self.rank = [0 for i in range(n)]
    

    def find(self, x):
        
        if self.root[x] == x:
            return x
    
        
        self.root[x] = self.find(self.root[x])

        return self.root[x]
    

    def union(self, a, b):
        

        
        self.root[b] = a
      

    
    def connected(self, a, b):
        return self.find(a) == self.find(b)
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