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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.
1function TreeNode(val, left, right) {
2 this.val = (val===undefined ? 0 : val)
3 this.left = (left===undefined ? null : left)
4 this.right = (right===undefined ? null : right)
5}
6
7const goodNodes = function(root) {
8 const dfs = (node, maxVal) => {
9 if (!node) return 0;
10 let good = 0;
11 if (node.val >= maxVal) good = 1;
12 maxVal = Math.max(maxVal, node.val);
13 good += dfs(node.left, maxVal);
14 good += dfs(node.right, maxVal);
15 return good;
16 };
17 return dfs(root, root.val);
18};
19
This JavaScript solution defines a DFS function inline and is called recursively with the maximum value found on the path as a parameter. The function calculates the number of good nodes during the traversal.
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.
using System.Collections.Generic;
public class TreeNode {
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
this.val = val;
this.left = left;
this.right = right;
}
}
public class Solution {
public int GoodNodes(TreeNode root) {
if (root == null) return 0;
Queue<(TreeNode, int)> queue = new Queue<(TreeNode, int)>();
queue.Enqueue((root, root.val));
int goodCount = 0;
while (queue.Count > 0) {
(TreeNode node, int maxVal) = queue.Dequeue();
if (node.val >= maxVal) goodCount++;
if (node.left != null) queue.Enqueue((node.left, Math.Max(maxVal, node.val)));
if (node.right != null) queue.Enqueue((node.right, Math.Max(maxVal, node.val)));
}
return goodCount;
}
}
In this C# solution, a queue is used to implement BFS, with each item being a tuple that contains both the current node and the maximum value encountered so far. The process checks each node, updating the count of good nodes as necessary.