<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 DisjointSet {  
public:
    vector<int> rank, parent, size;
    DisjointSet(int n) {
        rank.resize(n+1, 0); 
        parent.resize(n+1);
        size.resize(n+1); 
        for(int i = 0;i<=n;i++) {
            parent[i] = i; 
            size[i] = 1; 
        }
    }

    int findUPar(int node) {
        if(node == parent[node])
            return node; 
        return parent[node] = findUPar(parent[node]); 
    }

    void unionBySize(int u, int v) {
        int ulp_u = findUPar(u); 
        int ulp_v = findUPar(v); 
        if(ulp_u == ulp_v) return; 
        if(size[ulp_u] < size[ulp_v]) {
            parent[ulp_u] = ulp_v; 
            size[ulp_v] += size[ulp_u]; 
        }
        else {
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v]; 
        }
    }
};






class Solution {
public:
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
  // DSU approach
  // we will check the total number of components wo 1 hi hona chahiye hai
  // while making DSU agar already present hua and then humein again milega toh return false.
    
    DisjointSet ds(n);
    // take a map for counting the nodes and if it occurs more than once return false
    unordered_map<int,int> mpp;

    for(int i=0;i<n;i++)
    {
       
        if(leftChild[i]!=-1)
        {
        mpp[leftChild[i]]++;
        if(mpp[leftChild[i]]>1) return false;
            
            if(ds.findUPar(i)==ds.findUPar(leftChild[i])) return false;
            else
            {
                ds.unionBySize(i,leftChild[i]);
            }
        }

        if(rightChild[i]!=-1)
        {
             mpp[rightChild[i]]++;
           if(mpp[rightChild[i]]>1) return false;
            if(ds.findUPar(i)==ds.findUPar(rightChild[i])) return false;
            else
            {
                ds.unionBySize(i,rightChild[i]);
            }
        }
    }
    int cnt=0;
    for(int i=0;i<n;i++)
    {
        if(ds.parent[i]==i) cnt++;
    }
    if(cnt>1) return false;

    return true;

    }
};
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