题解 | #左叶子之和#简洁

代码简洁，递归核心是判断条件：左叶子之和，那只要找到左子节点是叶子节点的判断条件即可

c++实现

class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        // write code here
        if(!root) return 0;
        int sum_num=0;
      //判断条件：如果左子节点是叶子节点，则再加上右子树的左叶子之和即可
        if(root->left && !root->left->left && !root->left->right){  
            sum_num = root->left->val;
            sum_num += sumOfLeftLeaves(root->right);
            return sum_num;
        }else{  //如果左子节点不是叶子节点，就要将左右子树的左叶子之和相加
            sum_num += sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
            return sum_num;
        }
    }
};