<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 {
private:
    bool dfs(int node,vector<int>& leftChild,vector<int>& rightChild,vector<bool>& vis){
        if(node==-1) return true;
        if(vis[node]) return false;
        vis[node]=true;
        return dfs(leftChild[node],leftChild,rightChild,vis) && dfs(rightChild[node],leftChild,rightChild,vis);
    }
public:
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        vector<vector<int>> adj(n);
        vector<int> indegree(n);
        //if indegrree>1, then it's not a valid tree
        unordered_map<int,int> mp;
        for(auto it:leftChild) mp[it]++;
        for(auto it:rightChild){
            if(it!=-1 && mp.find(it)!=mp.end()) return false;
        }
        for(int i=0;i<n;i++){
            if(leftChild[i]!=-1 && indegree[leftChild[i]]++ ==1) return false;
            if(rightChild[i]!=-1 && indegree[rightChild[i]]++ ==1) return false;
        }
        //there should only be single root
        int root=-1;
        for(int i=0;i<n;i++){
            if(indegree[i]==0){
                if(root!=-1) return false; //can't have multiple roots
                root=i;
            }
        }
        vector<bool> vis(n,false);
        //all elements should be reachable
        bool res = (root!=-1) && dfs(root,leftChild,rightChild,vis);
        if(res){
            for(auto it:vis){
                if(!it) return false;
            }
            return true;
        }else return false;

    }
};
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