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

    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];
        }
    }
    int checkcomponent(int n){
    set<int> s;
    for(int i = 0; i < n; i++){
        s.insert(findUPar(i));
    }
    return s.size();
    } 
};
class Solution {
public:
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild,
                                 vector<int>& rightChild) {
        vector<int> indegree(n, 0);
    DisjointSet ds(n);
    for (int i = 0; i < n; i++) {
        if (leftChild[i] != -1) {
            indegree[leftChild[i]]++;
            if (indegree[leftChild[i]] > 1 || ds.findUPar(leftChild[i]) == ds.findUPar(i)) {
                return false;
            }
            ds.unionByRank(i, leftChild[i]);
        }
        if (rightChild[i] != -1) {
            indegree[rightChild[i]]++;
            if (indegree[rightChild[i]] > 1 || ds.findUPar(rightChild[i]) == ds.findUPar(i)) {
                return false;
            }
            ds.unionByRank(i, rightChild[i]);
        }
    }
    int rootCount = 0;
    for (int i = 0; i < n; i++) {
        if (indegree[i] == 0) {
            rootCount++;
        }
    }
    if (rootCount != 1) {
        return false;
    }

    return (ds.checkcomponent(n) == 1);
}
};
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