Given a binary tree root, a node X in the tree is named&nbsp;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.

&nbsp;
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 -&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.

Example 2:



Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -&gt; (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.

&nbsp;
Constraints:


	The number of nodes in the binary tree is in the range&nbsp;[1, 10^5].
	Each node's value is between [-10^4, 10^4].

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def goodNodes(self, root: TreeNode) -> int:
        
        def dfs(node,maxVal):
            if not node:
                return 0
            res=1 if node.val>=maxVal else 0
            maxVal=max(maxVal,node.val)
            res+=dfs(node.left,maxVal)
            res+=dfs(node.right,maxVal)
            return res
        return dfs(root,root.val)
class Solution {
    int ans = 0;
    public int goodNodes(TreeNode root) {
        if (root == null) return 0;
        dfs(root, root.val);
        return ans;
    }

    void dfs(TreeNode root, int mx) {
        if (root == null) return;

        mx = Math.max(mx, root.val);
        if(mx <= root.val) ans++;

        dfs(root.left, mx);
        dfs(root.right, mx);

    }
}
class Solution {
public:
    int count=0;
    void sol(TreeNode* root,int gr){
        if(!root)return;  // base condition 
       
        if(gr<=root->val){  //check point max element increase count
            gr=max(gr,root->val);
            count++;
        }
        
        if(root->left) sol(root->left,gr);
        if(root->right) sol(root->right,gr);
        
    }
    int goodNodes(TreeNode* root) {
        
        if(!root->left && !root->right) return 1; //check for if there is one node
        int gr=root->val;
        sol(root,gr);
        return count;
        
    }
};
var goodNodes = function(root) {
    let ans = 0;
    const traverse = (r = root, mx = root.val) => {
        if(!r) return;
        if(r.val >= mx) {
            ans++;
        }
        let childMax = Math.max(mx, r.val);
        traverse(r.left, childMax);
        traverse(r.right, childMax);
    }
    traverse();
    return ans;
};
Count Good Nodes in Binary Tree