12ms c++ 3 lines solution


  • 1
    T
    bool hasPathSum(TreeNode* root, int sum) {
        if(root == NULL) return false;
        if(root->left == NULL && root->right == NULL) return root->val == sum;
        return (hasPathSum(root->left, sum - root->val)+hasPathSum(root->right, sum - root->val));
    }

  • 0
    F

    The same as I did, but return root->val == sum is great!


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