<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:
    int countNode(int &root, vector<int> &leftChild, vector<int> &rightChild){
        queue<pair<int, int>> traversal;
        traversal.push({root, -1});
        int count = 0;
        while(!traversal.empty()){
            int size = traversal.size();
            while(size--){
                int node = traversal.front().first;
                int parent = traversal.front().second;
                count++;
                traversal.pop();

                if(leftChild[node] != -1 and leftChild[node] != parent)
                    traversal.push({leftChild[node], node});

                if(rightChild[node] != -1 and rightChild[node] != parent)
                    traversal.push({rightChild[node], node});
            }
        }
        // cout<<count;
        return count;
    }
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        map<int, int> parent;
        for(int i = 0; i<n; i++){
            if(leftChild[i] != -1){
                if(parent.find(leftChild[i]) != parent.end())
                    return false;
                else
                    parent[leftChild[i]] = i;
            }

            if(rightChild[i] != -1){
                if(parent.find(rightChild[i]) != parent.end())
                    return false;
                else
                    parent[rightChild[i]] = i;
            }
        }
        cout<<"x";
        int node = 0 ,root_node = -1;
        for(auto &x : parent){
            if(x.first != node){
                // if(root_node != -1) return false;
                root_node = node;
                node++;
            }
            node++;
        }
        if(node == n-1) root_node = node;
        cout<<"x";
        if(root_node == -1) return false;
        cout<<root_node;
        cout<<"x";
        return countNode(root_node, leftChild, rightChild) == n;
    }
};
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