<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 dfs(TreeNode* &root, vector<int> &leftChild, vector<int> &rightChild, unordered_map<int, int> &myMap, int &n) {
        if (root == NULL) return true;

        if (leftChild[root->val] != -1) {
            --n;
            myMap[leftChild[root->val]]++;
            if (myMap[leftChild[root->val]] > 1) return false;
            root->left = new TreeNode(leftChild[root->val]);
        }

        bool left = dfs(root->left, leftChild, rightChild, myMap, n);

        if (rightChild[root->val] != -1) {
            --n;
            myMap[rightChild[root->val]]++;
            if (myMap[rightChild[root->val]] > 1) return false;
            root->right = new TreeNode(rightChild[root->val]);
        }

        bool right = dfs(root->right, leftChild, rightChild, myMap, n);

        return right && left;
    }

    bool validateBinaryTreeNodes(int n, vector<int>& leftChild, vector<int>& rightChild) {
        int principalRoot;
        int suma = 0;
        int sumaVerify = 0;

        for (int i=0;i<n;++i) suma += i;

        for (int i=0;i<leftChild.size();++i) {
            if (leftChild[i] != -1) sumaVerify += leftChild[i];
            if (rightChild[i] != -1) sumaVerify += rightChild[i];
        }

        if (sumaVerify > suma) {
            principalRoot = 0;
        } else {
            principalRoot = suma-sumaVerify;
        }

        TreeNode* root = new TreeNode(principalRoot);
        unordered_map<int, int> myMap;
        int counter = n-1;
        myMap[principalRoot]++;


        bool result = dfs(root, leftChild, rightChild, myMap, counter);
        if (counter != 0) return false;
        return result;
    }
};
49047