<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) {
        vector<int> indegree(n,0);
        vector<int> adj[n];
        int sum = 0;
        int cnt = 0;
        unordered_map<int,int> parent;
        queue<int> q;
        vector<int> ans;
        for(int i=0;i<n;i++) {
            if(leftChild[i]!=-1) {
                if(parent.find(i)!=parent.end() and parent[i] == leftChild[i]) return 0;
                indegree[leftChild[i]]++;
                // indegree[i]--;
                parent[leftChild[i]] = i;
                adj[i].push_back(leftChild[i]);
            }
            if(rightChild[i]!=-1) {
                if(parent.find(i)!=parent.end() and parent[i] == rightChild[i]) return 0;
                indegree[rightChild[i]]++;
                // indegree[i]--;
                parent[rightChild[i]] = i;
                adj[i].push_back(rightChild[i]);
            }
        }
        for(int i=0;i<n;i++) {
            if(indegree[i]>1) return 0;
            if(parent.find(i) == parent.end()) cnt++;
        }
        for(int i=0;i<n;i++) {
            if(indegree[i] == 0) q.push(i);
        }
        if(q.empty()) return 0;
        while(!q.empty()) {
            int node = q.front();
            ans.push_back(node);
            q.pop();
            for(auto it:adj[node]) {
                indegree[it]--;
                if(indegree[it] == 0) q.push(it);
            }
        }
        if(ans.size()<n) return 0;
        if(cnt>1) return 0;
        return 1;
    }
};
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