LeetCode75-1448-Count-Good-Nodes-BT

🧠 思路总结

Q: 为什么要单独写一个 helper 方法？

因为题目函数 goodNodes(root) 只给一个参数，但 DFS 需要两个参数 (当前节点, 当前路径最大值)。

✅ 代码（Java）

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public int goodNodes(TreeNode root) {
        return dfs(root, root.val);
    }
    private int dfs(TreeNode node, int max) {
        if (node == null) {
            return 0;
        }
        int count = 0;
        if (node.val >= max) {
            count = 1;
        }
        int newMax = Math.max(max, node.val);
        count += dfs(node.left, newMax);
        count += dfs(node.right, newMax);
        return count;
    }
}