Count Good Nodes in Binary Tree
<p>Given a binary tree <code>root</code>, a node <em>X</em> in the tree is named&nbsp;<strong>good</strong> if in the path from root to <em>X</em> there are no nodes with a value <em>greater than</em> X.</p>

<p>Return the number of <strong>good</strong> nodes in the binary tree.</p>

<p>&nbsp;</p>

<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1448.Count%20Good%20Nodes%20in%20Binary%20Tree/images/test_sample_1.png" style="width: 263px; height: 156px;" /></strong></p>

<pre>

<strong>Input:</strong> root = [3,1,4,3,null,1,5]

<strong>Output:</strong> 4

<strong>Explanation:</strong> Nodes in blue are <strong>good</strong>.

Root Node (3) is always a good node.

Node 4 -&gt; (3,4) is the maximum value in the path starting from the root.

Node 5 -&gt; (3,4,5) is the maximum value in the path

Node 3 -&gt; (3,1,3) is the maximum value in the path.</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1400-1499/1448.Count%20Good%20Nodes%20in%20Binary%20Tree/images/test_sample_2.png" style="width: 157px; height: 161px;" /></strong></p>

<pre>

<strong>Input:</strong> root = [3,3,null,4,2]

<strong>Output:</strong> 3

<strong>Explanation:</strong> Node 2 -&gt; (3, 3, 2) is not good, because &quot;3&quot; is higher than it.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>

<strong>Input:</strong> root = [1]

<strong>Output:</strong> 1

<strong>Explanation:</strong> Root is considered as <strong>good</strong>.</pre>

<p>&nbsp;</p>

<p><strong>Constraints:</strong></p>

<ul>

    <li>The number of nodes in the binary tree is in the range&nbsp;<code>[1, 10^5]</code>.</li>

    <li>Each node&#39;s value is between <code>[-10^4, 10^4]</code>.</li>

</ul>
Tree; Depth-First Search; Breadth-First Search; Binary Tree
/**
 * 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 {
    private int ans = 0;

    public int goodNodes(TreeNode root) {
        dfs(root, -100000);
        return ans;
    }

    private void dfs(TreeNode root, int mx) {
        if (root == null) {
            return;
        }
        if (mx <= root.val) {
            ++ans;
            mx = root.val;
        }
        dfs(root.left, mx);
        dfs(root.right, mx);
    }
}