<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, lc: List[int], rc: List[int]) -> bool:
        if n == 1 and (lc[0] == 0 or rc[0] == 0): return False
        if n == 1: return True
        
        par = {}
        adj = [[] for i in range(n)]
        
        for i in range(n):
            l, r = lc[i], rc[i]
            
            if l != -1:
                adj[i].append(l)
                if l not in par: 
                    par[l] = 1
                else:
                    par[l] += 1
                    if par[l] >= 2: 
                        print("comm par l f")
                        return False

            if r != -1:
                adj[i].append(r)
                if r not in par: 
                    par[r] = 1
                else:
                    par[r] += 1
                    if par[r] >= 2:
                        print("comm par r f")
                        return False

        vis = [False] * n
        dfs_vis = [False] * n
        def detect_cycle(node: int) -> bool:
            vis[node] = True
            dfs_vis[node] = True
            
            for neigh in adj[node]:
                if vis[neigh] and dfs_vis[neigh]: return True
                
                if detect_cycle(neigh): return True
            
            dfs_vis[node] = False
            return False
        
        if not par: return False

        for i in range(n):
            if i in par: continue
            
            if detect_cycle(i): 
                return False
            break
        
        for v in vis:
            if not v:
                print("diff comps f")
                return False


        return True
        
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