<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, vector<int>> mp;
        vector<int> inDegree(n, 0);
        int i = 0;
        for (i = 0; i < n; ++i)
        {
            if (leftChild[i] != -1) 
            {
                mp[i].push_back(leftChild[i]);
                ++inDegree[leftChild[i]];
            }
            if (rightChild[i] != -1) 
            {
                mp[i].push_back(rightChild[i]);
                ++inDegree[rightChild[i]];
            }
        }

        // for (auto kvp : mp)
        // {
        //     cout << "key: " << kvp.first << endl;
        //     for (auto num : kvp.second) cout << num << " ";
        //     cout << endl;
        // }
        int root = -1;
        for (i = 0; i < n; ++i)
        {
            if (inDegree[i] != 0) continue;
            if (root != -1) return false;
            root = i;
        }

        if (root == -1) return false;

        stack<int> stk;
        stk.push(root);
        vector<bool> visited(n, false);
        while (!stk.empty())
        {
            int cur = stk.top(); stk.pop();
            visited[cur] = true;
            if (mp.count(cur))
            {
                for (auto n : mp[cur])
                {
                    if (visited[n]) return false;
                    stk.push(n);
                }
            }
        }

        for (auto v : visited)
        {
            if (!v) return false;
        }

        return true;
    }
};
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