938. Range Sum of BST 二叉树范围内求和

2021/7/19 6:05:57

本文主要是介绍938. Range Sum of BST 二叉树范围内求和,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

Given the root node of a binary search tree and two integers low and high, return the sum of values of all nodes with a value in the inclusive range [low, high].

 

Example 1:

Input: root = [10,5,15,3,7,null,18], low = 7, high = 15
Output: 32
Explanation: Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32.

Example 2:

Input: root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10
Output: 23
Explanation: Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23.

你好,剪枝法

/**
 * 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 rangeSumBST(TreeNode root, int L, int R) {
        if(root == null) return 0;
        if(root.val > R) return rangeSumBST(root.left, L, R);
        if(root.val < L) return rangeSumBST(root.right, L, R);
        return root.val + rangeSumBST(root.left, L, R) + rangeSumBST(root.right, L, R);      
    }
}

 


 



这篇关于938. Range Sum of BST 二叉树范围内求和的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!


扫一扫关注最新编程教程