Leetcode #1448
I’m super rusty with graph traversal, so I spent reallyyyyy long staring at this and overcomplicating it in my head. This turned out to be much easier than I expected.
Problem
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:
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:
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].
My Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
/*
DFS helper:
- input treenode and max`
- if node.val >= max, count++
- update max
- returns number of good nodes
+ dfs(left, newMax) + dfs(right, newMax)
- begin with dfs(root, root.val)
*/
class Solution {
public int goodNodes(TreeNode root) {
return DFS(root, root.val);
}
private int DFS(TreeNode node, int max) {
if (node == null) {
return 0;
}
int count = (node.val >= max) ? 1 : 0;
int newMax = Math.max(max, node.val);
return count + DFS(node.left, newMax) + DFS(node.right, newMax);
}
}