<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:
        adj_map = defaultdict(list)
        for i in range(n):
            if leftChild[i] != -1:
                adj_map[i].append(leftChild[i])
            if rightChild[i] != -1:
                adj_map[i].append(rightChild[i])
        visited = [0] * n
        indegrees = [0] * n
        for num in leftChild + rightChild:
            if num != -1:
                indegrees[num] += 1
        
        root = -1
        for i in range(len(indegrees)):
            if indegrees[i] == 0:
                root = i
        if root == -1:
            return False
        print(indegrees)
        print(root)
        is_acyclic = self.is_acyclic(root, visited, adj_map)
        print('is_acyclic:', is_acyclic)
        no_shared_parents = self.no_shared_parents(leftChild, rightChild)
        print('no_shared_parents:', no_shared_parents)
        is_connected = self.is_connected(root, n, adj_map)
        print('is_connected:', is_connected)
        return is_acyclic and no_shared_parents and is_connected


    def is_acyclic(self, node, visited, adj_map):
        if visited[node] == 1:
            return False
        visited[node] = 1
        for neighbor in adj_map[node]:
            if not self.is_acyclic(neighbor, visited, adj_map):
                return False
        return True

    def no_shared_parents(self, leftChild, rightChild):
        node_set = set()
        for num in leftChild + rightChild:
            if num != -1:
                if num in node_set:
                    return False
                node_set.add(num)
        return True
    
    def get_count_from_root(self, node, adj_map, visited):
        if visited[node] == 1:
            return 0
        visited[node] = 1
        count = 1
        for neighbor in adj_map[node]:
            count += self.get_count_from_root(neighbor, adj_map, visited)
        return count


    def is_connected(self, root, n, adj_map):
        visited = [0] * n
        count = self.get_count_from_root(root, adj_map, visited)
        return n == count

20970