<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>

// Approach 2: Breadth First Search (BFS)
class Solution {
public:
    int findRoot(int n, vector<int> leftChild, vector<int> rightChild) {
        unordered_set<int> children;
        children.insert(leftChild.begin(), leftChild.end());
        children.insert(rightChild.begin(), rightChild.end());
        for(int i = 0; i < n; i++) {
            if(children.find(i) == children.end()) {
                return i;
            }
        }
        return -1;
    }
    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        int root = findRoot(n, leftChild, rightChild);
        if(root == -1) {
            return false;
        }
        unordered_set<int> visited;
        queue<int> q;
        visited.insert(root);
        q.push(root);
        while(!q.empty()) {
            int curr = q.front();
            q.pop();
            int children[] = {leftChild[curr], rightChild[curr]};
            for(int child : children) {
                if(child != -1) {
                    if(visited.find(child) != visited.end()) {
                        return false;
                    }
                    q.push(child);
                    visited.insert(child);
                }
            }
        }
        return visited.size() == n;
    }
};
// Complexity Analysis
// Time complexity: O(n)
// To find the root, we convert leftChild and rightChild to a set, which costs O(n). Then, we iterate over all nodes, which also costs O(n).
// Once we have the root, we perform a BFS that costs O(n) as we never visit a node more than once. Note that an efficient queue implementation with O(1) operations is required to achieve this complexity.
// Space complexity: O(n)
// We require O(n) space when converting leftChild and rightChild to a set to find the root. We also require O(n) space for queue and seen during the BFS.
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