Count Nodes That Match or Exceed Every Ancestor

### Problem A native binary-tree node is good when no value on its root-to-node path is greater than its own value. Return the number of good nodes, counting the root and allowing equality with ancestors. ```cpp class Solution { public: int goodNodes(TreeNode* root); }; ``` ### Examples **Example 1:** ```text Input: root = [2,2,1,null,3] Output: 3 Explanation: The root2,the left child2,and its child3 are good. The right child1 is below its ancestor2. ``` **Example 2:** ```text Input: root = [-2,-3,-1] Output: 2 Explanation: The root-2 and right child-1 qualify; left child-3 does not. ``` ### Constraints - The binary tree contains 1 to 100000 nodes. - Every value lies in `[-10000,10000]`. ### Contract considerations Preserve the TreeNode pointer interface; level-order arrays describe the examples only. ```hint Read the boundary condition Compare a node against its full root-to-node path, allowing equality. Include negative values and the root itself. ``` ### Boundary reminders Preserve the TreeNode pointer interface; level-order arrays describe the examples only.