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？？
