222. 完全二叉树的节点个数

通过的代码：

/*** 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 geNum(TreeNode* cur){//确定递归函数参数和返回值if(cur==nullptr) return 0;//确定终止条件int leftNum=0,rightNum=0;TreeNode* left=cur->left;TreeNode* right=cur->right;while(left){leftNum++;left=left->left;}while(right){rightNum++;right=right->right;}if(leftNum==rightNum) return (2<<leftNum)-1;//单层递归逻辑int leftTreeNum=geNum(cur->left);int rightTreeNum=geNum(cur->right);int midTreeNum=leftTreeNum+rightTreeNum+1;return midTreeNum;}int countNodes(TreeNode* root) {return geNum(root);}
};

未通过的代码：

/*** 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 geNum(TreeNode* cur){//确定递归函数参数和返回值//if(cur==nullptr) return 0;//确定终止条件int leftNum=0,rightNum=0;TreeNode* left=cur->left;TreeNode* right=cur->right;while(left){leftNum++;left=left->left;}while(right){rightNum++;right=right->right;}if(leftNum==rightNum) return (2<<leftNum)-1;//单层递归逻辑int leftTreeNum=geNum(cur->left);int rightTreeNum=geNum(cur->right);int midTreeNum=leftTreeNum+rightTreeNum+1;return midTreeNum;}int countNodes(TreeNode* root) {if(root==nullptr) return 0;return geNum(root);}
};

问题：

在countNodes函数中添加的代码:

if(root==nullptr) return 0;

好像没有用似的，必须在递归函数中添加这行代码才不会报错。

这是为什么？？