1448 - Count Good Nodes in Binary Tree (Medium)

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 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path
Node 3 -> (3,1,3) is the maximum value in the path.

Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.

Input: root = [1]
Output: 1
Explanation: Root is considered as good.

// Time complexity: O(N) as we visit every node only once.
// Space complexity: O(H) where H is the height of the tree.
// In the worst case, H would be N given that the tree only has one path.
class Solution {
public:
    // the idea is to record the max value from the root to the node
    // we can first initialise mx as INT_MIN
    int goodNodes(TreeNode* root, int mx = INT_MIN) {
        // if the root is null, we return 0
        if (!root) return 0;
        // then we can break it into 3 parts
        // the first part is the current node
        // if the current node value is greater than the maximum value so far
        // that means the current node is a good node
        // hence we add 1, else add 0
        return (mx <= root->val ? 1 : 0) +
                // the second part is the result of the left sub-tree
                // we traverse it with the updated maximum value at the current point
                // we don't need to check if root->left is null or not here
                // as we cover the null case in the first line
                goodNodes(root->left, max(root->val, mx)) +
                // the last part is the result of the right sub-tree
                // we traverse it with the updated maximum value at the current point
                // we don't need to check if root->right is null or not here
                // as we cover the null case in the first line
                goodNodes(root->right, max(root->val, mx));
    }
};

# 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:
    # Iterative Depth-First Search
    # We can simply solve this problem by running a DFS, and passing down a max
    # value to each node. A node is only good of it is >= to this max value.
    # Time Complexity: O(n). We are going to have to precess each node.
    # Space Complexity: O(h). Using a stack, we only ever have the current path
    # inside our stack.
    def goodNodes(self, root: TreeNode) -> int:
        # Early termination
        if not root:
            return 0
        # initialize return integer
        num_good_nodes = 0
        # initialize our stack, each item will be a tuple (node, max_value)
        # where the max value is the max value we have seen thus far going down
        # the path to the node.
        stack = [(root, root.val)]
        # while our stack still has nodes inside.
        while stack:
            # pop off the node and the highest value we have seen thus far.
            node, high = stack.pop()
            # Increment our num_good_nodes integer iff our current nodes value
            # is greater than or equal to the highest node we have seen.
            if node.val >= high:
                num_good_nodes += 1
            # if node has a left child
            if node.left:
                # add left child to the stack with the highest value.
                # highest value will be larger of the current nodes value and
                # the highest value that was passed to this iteration.
                stack.append((node.left, max(node.val,high)))
            # if the node has a right child
            if node.right:
                # add right child, make sure to update the high value.
                stack.append((node.right, max(node.val, high)))
        # return the number of good nodse that we calculated above.
        return num_good_nodes