<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) {
        vector<int> inDegree(n, 0);

        for (int i = 0; i < n; i++) {
            if (leftChild[i] != -1) {
                inDegree[leftChild[i]]++;
                if (inDegree[leftChild[i]] > 1) {
                    return false; // More than one incoming edge to a node
                }
            }

            if (rightChild[i] != -1) {
                inDegree[rightChild[i]]++;
                if (inDegree[rightChild[i]] > 1) {
                    return false; // More than one incoming edge to a node
                }
            }
        }

        int rootCount = 0;
        int rootCandidate = -1;
        for (int i = 0; i < n; i++) {
            if (inDegree[i] == 0) {
                rootCount++;
                if (rootCount > 1) {
                    return false; // More than one root node
                }
                rootCandidate = i;
            }
        }

        if (rootCount == 0) {
            return false; // No root node found
        }

        if (rootCount == 1) {
            return isConnected(n, leftChild, rightChild, rootCandidate);
        }

        return false;
    }

    bool isConnected(int n, vector<int>& leftChild, vector<int>& rightChild, int root) {
        vector<bool> visited(n, false);
        stack<int> dfs;
        dfs.push(root);

        while (!dfs.empty()) {
            int node = dfs.top();
            dfs.pop();

            if (visited[node]) {
                return false; // Cycle detected
            }

            visited[node] = true;

            if (leftChild[node] != -1) {
                dfs.push(leftChild[node]);
            }
            if (rightChild[node] != -1) {
                dfs.push(rightChild[node]);
            }
        }

        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                return false; // Some nodes are not reachable
            }
        }

        return true;
    }
};
34872