题解 | #左叶子之和#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return int整型
     */
    int sumOfLeftLeaves(TreeNode* root) {
        // write code here
        if(root == nullptr)
            return 0;
        int sum = 0;
        if(root->left && root->left->left == nullptr && root->left->right == nullptr)	//对应到当前节点有左叶子时，返回左叶子的值即可。
            sum = root->left->val;
        int left = sumOfLeftLeaves(root->left);
        
        int right = sumOfLeftLeaves(root->right);
        
        sum += left + right;
        return sum;
        
    }
};