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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};
19This 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.
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.