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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.
1class TreeNode {
2 int val;
3 TreeNode left;
4 TreeNode right;
5 TreeNode() {}
6 TreeNode(int val) { this.val = val; }
7 TreeNode(int val, TreeNode left, TreeNode right) {
8 this.val = val;
9 this.left = left;
10 this.right = right;
11 }
12}
13
14class Solution {
15 private int dfs(TreeNode node, int maxVal) {
16 if (node == null) return 0;
17 int count = 0;
18 if (node.val >= maxVal) count = 1;
19 maxVal = Math.max(maxVal, node.val);
20 count += dfs(node.left, maxVal);
21 count += dfs(node.right, maxVal);
22 return count;
23 }
24
25 public int goodNodes(TreeNode root) {
26 return dfs(root, root.val);
27 }
28}
29This Java implementation uses a helper method dfs encapsulated in the Solution class, which recursively traverses the binary tree and counts good nodes.
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.
1
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.