1448 - Count Good Nodes in Binary Tree (Medium)
Problem Statement
Given a binary tree root, a node X in the tree is named 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.
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 -> (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.
Example 2:
Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -> (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.
Constraints:
- The number of nodes in the binary tree is in the range
[1, 10^5]. - Each node's value is between
[-10^4, 10^4].
Approach 1: DFS
- C++
- Python
// 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