<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:
        dic = {}
        dic_child = {}
        for i in range(n):
            dic[i] = []
            dic_child[i] = []
        for i in range(n):
            if leftChild[i] >= 0:
                dic[leftChild[i]].append(i)
                dic_child[i].append(leftChild[i])
            if rightChild[i] >= 0:
                dic[rightChild[i]].append(i)
                dic_child[i].append(rightChild[i])

        root = 0
        visited = set()
        visited.add(root)
        while len(dic[root]) > 0:
            if dic[root][0] not in visited:
                visited.add(dic[root][0])
                root = dic[root][0]
            else:
                break

        self.dic_child = dic_child
        self.visited = set()
        self.valid = True

        self.dfs(root)
        return self.valid and len(self.visited) == n

    def dfs(self, node):
        if node in self.visited:
            self.valid = False
            return
        self.visited.add(node)
        for neighbor in self.dic_child[node]:
            if neighbor in self.visited:
                self.valid = False
                return
            else:
                self.dfs(neighbor)



        


        # count_no_parent = 0
        # for k, v in dic.items():
        #     if len(v) > 1:
        #         return False
        #     if len(v) == 0:
        #         count_no_parent += 1
        # return True if count_no_parent == 1 else False
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