Problem: 1448. 统计二叉树中好节点的数目

题目描述

思路

对二叉树进行先序遍历，边遍历边对比并更新当前路径上的最大值pathMax，若当pathMax小于等于当前节点值，则好节点的数目加一

复杂度

时间复杂度:

$O(n)$;其中$n$为二叉树的节点个数

空间复杂度:

$O(h)$；其中$h$为二叉树的高度

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 = 0;
    public int goodNodes(TreeNode root) {
    
      
        traverse(root, root.val);
        return count;
    }

    private void traverse(TreeNode root, int pathMax) {
    
      
        if (root == null) {
    
      
            return;
        }
        // Find a "good node"
        if (pathMax <= root.val) {
    
      
            count++;
            // Update the maximum value on the path
            pathMax = Math.max(root.val, pathMax);
        }
        traverse(root.left, pathMax);
        traverse(root.right, pathMax);
    }
}