LeetCode 222. Count Complete Tree Nodes

https://leetcode.com/problems/count-complete-tree-nodes/description/
简单的二分法分治，探测（子）树的左右侧是否相等即可。
（下面代码没实现）想要更快的话，其实自从第一次算出左侧的深度后，后面的左测深度直接可以通过减法求出，不用再去探测一次了。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int countLevel(TreeNode* root, bool to_left) {
        int count = 0;
        while (root != nullptr)
        {
            count++;
            root = to_left ? root->left : root->right;
        }
        return count;
    }
    int countNodes(TreeNode* root) {
        if (root == nullptr)
            return 0;
        int lvl_left = countLevel(root, true);
        int lvl_right = countLevel(root, false);
        if (lvl_left == lvl_right)
        {
            // finally
            return (1 << lvl_left) - 1; // 2^n - 1
        }
        else
        {
            return 1 + countNodes(root->left) + countNodes(root->right);
        }
    }
};