LeetCode每日一题(669. Trim a Binary Search Tree)

这篇博客讨论了如何修剪一个二叉搜索树，使其所有元素位于给定的low和high范围内。算法首先检查当前节点的值，如果小于low，则返回右子树；如果大于high，则返回左子树；如果在范围内，则递归处理左右子树。这种方法保留了原有的相对结构。给出了两个示例解释了该过程，并提供了具体的代码实现。

Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node’s descendant should remain a descendant). It can be proven that there is a unique answer.

Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.

Example 1:

Input: root = [1,0,2], low = 1, high = 2
Output: [1,null,2]

Example 2:

Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
Output: [3,2,null,1]

Constraints:

The number of nodes in the tree in the range [1, 104].
0 <= Node.val <= 104
The value of each node in the tree is unique.
root is guaranteed to be a valid binary search tree.
0 <= low <= high <= 104

二叉搜索树的特性是，左边所有的子节点都小于当前节点，右边所有子节点都大于当前节点。如果current_value < low则左侧所有子节点的值肯定也都< low, 这时候只需要返回右侧子节点。如果current_value > high则右侧所有子节点的值肯定也都> high, 这时只需要返回左侧子节点。如果low <= current_value <= high则递归调用处理左右两侧子节点

use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn trim_bst(
        root: Option<Rc<RefCell<TreeNode>>>,
        low: i32,
        high: i32,
    ) -> Option<Rc<RefCell<TreeNode>>> {
        if let Some(node) = root {
            if node.borrow().val < low {
                return Solution::trim_bst(node.borrow_mut().right.take(), low, high);
            }
            if node.borrow().val > high {
                return Solution::trim_bst(node.borrow_mut().left.take(), low, high);
            }
            let left = node.borrow_mut().left.take();
            let right = node.borrow_mut().right.take();
            node.borrow_mut().left = Solution::trim_bst(left, low, high);
            node.borrow_mut().right = Solution::trim_bst(right, low, high);
            return Some(node);
        }
        None
    }
}