<p>Given the <code>root</code> of a <strong>complete</strong> binary tree, return the number of the nodes in the tree.</p>

<p>According to <strong><a href="http://en.wikipedia.org/wiki/Binary_tree#Types_of_binary_trees" target="_blank">Wikipedia</a></strong>, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between <code>1</code> and <code>2<sup>h</sup></code> nodes inclusive at the last level <code>h</code>.</p>

<p>Design an algorithm that runs in less than&nbsp;<code data-stringify-type="code">O(n)</code>&nbsp;time complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/14/complete.jpg" style="width: 372px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6]
<strong>Output:</strong> 6
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = []
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 5 * 10<sup>4</sup>]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 5 * 10<sup>4</sup></code></li>
	<li>The tree is guaranteed to be <strong>complete</strong>.</li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
    static inline auto _ = [] {
        std::ios::sync_with_stdio(false);
        std::cin.tie(nullptr);
        std::ofstream out("user.out", std::ios::out | std::ios::binary);
        out.rdbuf()->pubsetbuf(nullptr, 256);
        std::string s;
        std::noskipws(std::cin);
        while (std::getline(std::cin, s)) {
            int count = 0;
            bool inNum = false;
            for (char c : s) {
                if (std::isdigit(c) != 0) {
                    if (!inNum) {
                        ++count;
                        inNum = true;
                    }
                } else {
                    inNum = false;
                }
            }
            out << count << '\n';
        }
        std::skipws(std::cin);
        out.flush();
        exit(0);
        return 0;
    }();

public:
    int countNodes(TreeNode* root) {

        if (root == NULL) {
            return 0;
        }

        int sum = 0;

        sum += countNodes(root->left) + 1;
        sum += countNodes(root->right);

        return sum;
    }
};
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