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Time Complexity: O(n), where n is the number of nodes in the binary tree.
Space Complexity: O(h), where h is the height of the tree due to the recursion stack.
1#include <iostream>
2using namespace std;
3
4struct TreeNode {
5 int val;
6 TreeNode *left;
7 TreeNode *right;
8 TreeNode() : val(0), left(nullptr), right(nullptr) {}
9 TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
10 TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
11};
12
13class Solution {
14public:
15 int dfs(TreeNode* node, int maxVal) {
16 if (!node) return 0;
17 int good = 0;
18 if (node->val >= maxVal) good = 1;
19 maxVal = max(maxVal, node->val);
20 return good + dfs(node->left, maxVal) + dfs(node->right, maxVal);
21 }
22
23 int goodNodes(TreeNode* root) {
24 return dfs(root, root->val);
25 }
26};
27This C++ solution uses a member function dfs of the Solution class to recursively traverse the tree and count the good nodes based on the maximum value constraint.
Time Complexity: O(n), where n is the number of nodes in the tree.
Space Complexity: O(w), where w is the maximum width of the tree.
The C solution implements an iterative BFS using a custom queue structure. Each item in the queue stores a node and the maximum value along the path from the root to that node. The algorithm iterates over the queue, evaluating each node's value against the max value, updating good nodes count accordingly.