1400 - 1499
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
// 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));
}
};1443 - Minimum Time to Collect All Apples in a Tree (Medium)
Author: @hirotake111, @wkw | https://leetcode.com/problems/minimum-time-to-collect-all-apples-in-a-tree/
1457 - Pseudo-Palindromic Paths in a Binary Tree (Medium)
Author: @wkw | https://leetcode.com/problems/pseudo-palindromic-paths-in-a-binary-tree/