<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 {
public:
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        
        unordered_map<int, list<int>> adjList;
        unordered_map<int, int> childParentMap;
        for(int i = 0; i < n; ++i)
        {
            int root = i;
            int left = leftChild[i];
            int right = rightChild[i];

            if(left != -1)
            {
                adjList[i].push_back(left);
                if(childParentMap.find(left) != childParentMap.end())
                    return false;
                childParentMap[left] = i;
            }
            if(right != -1)
            {
                adjList[i].push_back(right);
                if(childParentMap.find(right) != childParentMap.end())
                    return false;
                childParentMap[right] = i;
            }
        }

        int root = -1;
        for(int i = 0; i < n; ++i)
        {
            if(childParentMap.find(i) == childParentMap.end() )
            {
                if(root != -1)
                    return false;
                root = i;
            }
        }

        //if no root found
        if(root == -1)
            return false;

        //check if all nodes can be visited or not
        vector<bool> visited(n, false);
        visited[root] = true;
        queue<int> nodes;
        nodes.push(root);
        int cnt = 1;
        while(!nodes.empty())
        {
            int currNode = nodes.front();
            nodes.pop();
            for(int &neighbour : adjList[currNode])
            {
                if(!visited[neighbour])
                {
                    cnt++;
                    nodes.push(neighbour);
                    visited[neighbour] = true;
                }
            }
        }
        return cnt == n;
    }
};
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