<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, left_children_list, right_children_list) -> bool:
        memo_parent = {}
        
        memo_left_children = {}
        for i in range(0, len(left_children_list)):
            left_child = left_children_list[i]
            if (left_child != -1):
                if (left_child in memo_left_children.keys() and memo_left_children[left_child] == i):
                    return False
                memo_parent[left_child] = i
                memo_left_children[i] = left_child

        memo_right_children = {}
        for i in range(0, len(right_children_list)):
            right_child = right_children_list[i]
            if (right_child != -1):
                if (right_child in memo_right_children.keys() and memo_right_children[right_child] == i):
                    return False
                memo_parent[right_child] = i
                memo_right_children[i] = right_child

        for i in range(n-1, -1, -1):
            if (self.is_any_cycle(i, memo_parent)):
                return False

        children_intersection = set(memo_left_children.values()).intersection(set(memo_right_children.values()))
        if (len(children_intersection) > 0):
            return False
        
        children_set = set(memo_left_children.values()).union(set(memo_right_children.values()))
        if (len(set(range(0, n)).difference(children_set)) > 1):
            return False

        return True
    
    def is_any_cycle(self, i, memo_parent):
        curr_node = i
        seen = {}
        while (curr_node is not None):
            if (curr_node in seen.keys()):
                return True
            seen[curr_node] = True
            if (curr_node in memo_parent.keys()):
                curr_node = memo_parent[curr_node]
            else:
                curr_node = None
        
        return False
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