2022-09-01 Daily-Challenge

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In: DailyChallenge

Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 1

Description

Count Good Nodes in Binary Tree

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:

img

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:

img

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].

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  int goodNodes(TreeNode* root, int X = INT_MIN) {
    if(!root) return 0;
    if(root->val >= X) {
      return 1 + goodNodes(root->left, root->val) + goodNodes(root->right, root->val);
    } else {
      return goodNodes(root->left, X) + goodNodes(root->right, X);
    }
  }
};

// Accepted
// 63/63 cases passed (96 ms)
// Your runtime beats 99.23 % of cpp submissions
// Your memory usage beats 6.64 % of cpp submissions (86.6 MB)
Algorithm