1448. Count Good Nodes in Binary Tree

Question

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.

Solution

DFS搜索，每次传入当前分支的最大值。
如果当前值大于等于最大值，则count+1。

Code

/**
 * 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 {
    int count;
    public int goodNodes(TreeNode root) {
        count = 0;
        dfs(root, Integer.MIN_VALUE);
        return count;
    }
    
    private void dfs(TreeNode root, int max){
        if(root == null) return;
        if(root.val >= max){
            count++;
            max = root.val;
        }

        dfs(root.left, max);
        dfs(root.right, max);
    }
}