递归解法

class Solution {
public:
    bool hasPathSum(TreeNode* root, int targetSum) {
        if (!root) return false;

        //已经抵达叶子节点，判断targetSum是否等于叶子节点的值即可
        if (!root->left && !root->right) return root->val == targetSum;

        //递归遍历左右子节点，目标和减去当前节点的值
        return hasPathSum(root->left, targetSum - root->val) || hasPathSum(root->right, targetSum - root->val);
    }
};