1448. Count Good Nodes in Binary Tree

1. Description

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.

2. Example

Example 1

Input: root = [3,1,4,3,null,1,5]
Output: 4
Explanation: Nodes in blue are good.
Root Node (3) is always a good node.
Node 4 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path
Node 3 -> (3,1,3) is the maximum value in the path.

Example 2

Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -> (3, 3, 2) is not good, because “3” is higher than it.

Example 3

Input: root = [1]
Output: 1
Explanation: Root is considered as good.

3. Constraints

• The number of nodes in the binary tree is in the range [1, $10^5$].
• Each node’s value is between [-$10^4$, $10^4$].

4. Solutions

Depth-First Search

n is the number of nodes in root
Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    int goodNodes(TreeNode *root) {
        return count_good_nodes(root, INT_MIN);
    }

private:
    int count_good_nodes(TreeNode *root, int max_value) {
        if (root == nullptr) {
            return 0;
        }

        int child_max_value = max(root->val, max_value);
        return (root->val >= max_value ? 1 : 0) + count_good_nodes(root->left, child_max_value) +
            count_good_nodes(root->right, child_max_value);
    }
};