题解 | 二叉树中和为某一值的路径(一)

class Solution {
public:
    int ans = 0;

    bool hasPathSum(TreeNode* root, int sum) {
        if (!root) return false;
        ans += root->val;
        if (ans == sum && !root->left && !root->right) return true;

        bool l_tag = !root->left ? false: hasPathSum(root->left, sum), r_tag = !root->right ? false: hasPathSum(root->right, sum);
        ans -= root->val;
        return l_tag || r_tag;
    }
};