LeetCode 1448 Count Good Nodes in Binary Tree - Implementation

$dfs(node, pathMax)$: returns count of good nodes in subtree. Add 1 if $node.val \geq pathMax$. Recurse on children with $newMax = \max(pathMax, node.val)$. Sum the results. Base case: null returns $0$. Leaf returns $1$ if good, $0$ otherwise. Time: $O(n)$, Space: $O(h)$. This is the template for top-down tree DP: carry state down, aggregate results up. This is one of the simpler tree DP problems. It demonstrates the top-down pattern clearly.

Time complexity: $O(n)$.

Space complexity: $O(h)$.