[LeetCode] 1448. Count Good Nodes in Binary Tree

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].

统计二叉树中好节点的数目。

给你一棵根为 root 的二叉树，请你返回二叉树中好节点的数目。

「好节点」X 定义为：从根到该节点 X 所经过的节点中，没有任何节点的值大于 X 的值。

因为题目定义的路径是从根节点到叶子节点，所以思路天然是前序遍历。我们需要一个全局变量记录到底有多少节点是好节点。前序遍历的 helper 函数里也需要记录一个当前遇到的最大值 max，这样往子树递归的时候才知道当前路径上遇到的最大值是多少。

时间O(n)

空间O(n)

Java实现

/**
 * 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 res;
    
    public int goodNodes(TreeNode root) {
        res = 0;
        helper(root, Integer.MIN_VALUE);
        return res;
    }
    
    private void helper(TreeNode root, int max) {
        if (root == null) {
            return;
        }
        if (root.val >= max) {
            max = Math.max(max, root.val);
            res++;
        }
        helper(root.left, max);
        helper(root.right, max);
    }
}

 

LeetCode 题目总结