leetcode-938-Range-Sum-of-BST

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描述

Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).

The binary search tree is guaranteed to have unique values.

Example 1:

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Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
Output: 32

Example 2:

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Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
Output: 23

Note:

  1. The number of nodes in the tree is at most 10000.
  2. The final answer is guaranteed to be less than 2^31.

分析

给定一颗二叉树，以及 L，R，求值在 L，R 范围内的节点的值的和。

解决方案1(Python)

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def rangeSumBST(self, root, L, R):
        """
        :type root: TreeNode
        :type L: int
        :type R: int
        :rtype: int
        """
        if root == None:
            return 0
        now_item = root.val if L <= root.val and root.val <= R else 0
        return self.rangeSumBST(root.left, L, R) + self.rangeSumBST(root.right, L, R) + now_item

解决方案2(Python)

剪个枝，风味更佳。

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def rangeSumBST(self, root, L, R):
        """
        :type root: TreeNode
        :type L: int
        :type R: int
        :rtype: int
        """
        if root == None:
            return 0
        sum = 0
        if L <= root.val and root.val <= R:
            sum += root.val
        if root.val > L:
            sum += self.rangeSumBST(root.left, L, R)
        if root.val < R:
            sum += self.rangeSumBST(root.right, L, R)
        return sum

解决方案3(Rust)

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// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
// 
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    pub fn range_sum_bst(root: Option<Rc<RefCell<TreeNode>>>, l: i32, r: i32) -> i32 {
        if let Some(root) = root {
            let mut sum = 0;
            if l <= root.borrow().val && root.borrow().val <= r {
                sum += root.borrow().val;
            }
            
            if l < root.borrow().val {
                sum += Self::range_sum_bst(root.borrow().left.clone(), l, r);
            }
            if r > root.borrow().val {
                sum += Self::range_sum_bst(root.borrow().right.clone(), l, r);
            }
            return sum;
        }
        return 0;
    }
}

题目来源