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.
 * 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 {
public:
    int goodNodes(TreeNode* root) {
        int ans = 0;
        function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int mx) {
            if (!root) {
                return;
            }
            if (mx <= root->val) {
                ++ans;
                mx = root->val;
            }
            dfs(root->left, mx);
            dfs(root->right, mx);
        };
        dfs(root, -1e6);
        return ans;
    }
};