<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) {
        //adjacency list traverse ;
        unordered_map<int,vector<int>> adj;
        //child to parent store;
        //i. e. to identify who the parent of current child and make necessary validation
        unordered_map<int,int>child_to_parent;

        for(int i=0;i<n;i++){
            int node = i;
            int leftc = leftChild[i];
            int rightc = rightChild[i];
            if(leftc!=-1){
                adj[node].push_back(leftc);
                if(child_to_parent.find(leftc) != child_to_parent.end()){
                    return false;
                }else{
                    child_to_parent[leftc] = node;
                }
            }
            if(rightc!=-1){
                adj[node].push_back(rightc);
                if(child_to_parent.find(rightc) != child_to_parent.end()){
                    return false;
                }else{
                    child_to_parent[rightc] = node;
                }
            }  
        }

        int root = -1;
        for(int i = 0 ; i<n; i++){
            if(child_to_parent.find(i) == child_to_parent.end()){
                if(root != -1){
                    return false;
                }
                else{
                    root = i;
                }
            }
        }
        if(root == -1)return false;

        //bfs traversal for counting nodes';
        queue<int> q;
     
        vector<bool>visited(n,false);
        q.push(root);
        int count = 1;
        visited[root] = true;
        while(!q.empty()){
            int u = q.front();
            q.pop();

            for(auto v : adj[u]){
                if(!visited[v]){
                    visited[v] = true;
                    q.push(v);
                    count++;
                }
            }
        }
        if(count!=n)return false;

        return true;
    }
};
49362