代码随想录：左叶子之和

代码随想录：左叶子之和

没啥意思，很铸币的判断方法

/*** 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 sumOfLeftLeaves(TreeNode* root) { return tra(root); }int tra(TreeNode* node) {if (!node)return 0;int sum = 0;if (node->left) {if (node->left->left == NULL && node->left->right == NULL) {sum += node->left->val;} else {sum += tra(node->left);}}if (node->right) {sum += tra(node->right);}return sum;}
};