LeetCode-in-Java

1448. Count Good Nodes in Binary Tree

Medium

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:

Solution

import com_github_leetcode.TreeNode;

/*
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private int count = 0;

    private void traverse(TreeNode root, int max) {
        if (root == null) {
            return;
        }
        if (root.val >= max) {
            count += 1;
            max = root.val;
        }
        traverse(root.left, max);
        traverse(root.right, max);
    }

    public int goodNodes(TreeNode root) {
        traverse(root, Integer.MIN_VALUE);
        return count;
    }
}