Note：
就是递归的去求左子树节点的数量+右子树节点的数量+根节点
求当前子树的节点数量，就是先看看是不是满二叉树，如果是满二叉树，就直接求，否则就继续递归下去
代码如下：
/**
* 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 {
public:
int countNodes(TreeNode* root) {
if(!root) return 0;
auto l = root -> left, r = root -> right;
int x = 1, y = 1;
while(l) l = l -> left, x ++;
while(r) r = r -> right, y ++;
if(x == y) return (1 << x) - 1;
return countNodes(root -> left) + 1 + countNodes(root -> right);
}
};
