1448. Count Good Nodes in Binary Tree

Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.

Return the number of good nodes in the binary tree.

/**
 * 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 {
    int search(TreeNode* node, int X) {
        int res = node->val >= X;
        if(node->left != nullptr) {
            res += search(node->left, max(X, node->val));
        }
        if(node->right != nullptr) {
            res += search(node->right, max(X, node->val));
        }

        return res;
    }
public:
    int goodNodes(TreeNode* root) {
        return search(root, root->val);
    }
};

/**
 * 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 goodNodes(TreeNode* root, int X = INT_MIN) {
        return root == nullptr ? 0 : 
        (root->val >= X) + goodNodes(root->left, max(X, root->val)) + goodNodes(root->right, max(X, root->val));
    }
};