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1448. Count Good Nodes in Binary Tree

🔗 Source

Approach

Recursion, using variable max to log the previous maximum value and compare max with current root value.

Complexity

  • Time complexity: \(O(N)\)
  • Space complexity: \(O(1)\)

Code

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/**
 * 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;
 *     }
 * }
 */
class Solution {
    public int goodNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int max= root.val;
        return 1 + helper(root.left, max) + helper(root.right, max);
    }

    public int helper(TreeNode root, int max) {
        if (root == null) {
            return 0;
        }
        if (max > root.val) {
            return helper(root.left, max) + helper(root.right, max);
        }
        max = Math.max(max, root.val);
        return 1 + helper(root.left, max) + helper(root.right, max);
    }
}