404. 左叶子之和

题目链接

/**
 * 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) {
        if (!root) return 0;
        int mid = 0;
        int left = sumOfLeftLeaves(root->left);
        int right = sumOfLeftLeaves(root->right);
        if (root->left && !root->left->left && !root->left->right) {
            mid = root->left->val;
        }
        int sum = mid + left + right;
        return sum;
    }
};

总结

• 求左子叶子节点的和，不能直接使用层序遍历
• 左叶子节点需要初始化0，不然答案错误。