Count Good Nodes in Binary Tree

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.

 

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.  

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].

Code

Time Complexity: $O(n)$, Space Complexity: $O(1)$

/**
 * 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 res = 0;
public:
    int goodNodes(TreeNode* root) {
        if(!root) return 0;
        int curMax = INT_MIN;
        check(root, curMax);
        return res;
    }
 
    void check(TreeNode* node, int curMax) {
        if(!node) return;
        if(node->val >= curMax) {
            res++;
        }
        curMax = max(curMax, node->val);
        check(node->left, curMax);
        check(node->right, curMax);
    }
};

Source

• Count Good Nodes in Binary Tree - LeetCode