0669. Trim a Binary Search Tree

669. Trim a Binary Search Tree#

题目 #

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]
``````

Example 3:

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

Example 4:

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

Example 5:

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

Constraints:

• The number of nodes in the tree in the range `[1, 10^4]`.
• `0 <= Node.val <= 10^4`
• The value of each node in the tree is unique.
• `root` is guaranteed to be a valid binary search tree.
• `0 <= low <= high <= 10^4`

解题思路 #

• 这一题考察二叉搜索树中的递归遍历。递归遍历二叉搜索树每个结点，根据有序性，当前结点如果比 high 大，那么当前结点的右子树全部修剪掉，再递归修剪左子树；当前结点如果比 low 小，那么当前结点的左子树全部修剪掉，再递归修剪右子树。处理完越界的情况，剩下的情况都在区间内，分别递归修剪左子树和右子树即可。

代码 #

``````package leetcode

import (
"github.com/halfrost/leetcode-go/structures"
)

// TreeNode define
type TreeNode = structures.TreeNode

/**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/

func trimBST(root *TreeNode, low int, high int) *TreeNode {
if root == nil {
return root
}
if root.Val > high {
return trimBST(root.Left, low, high)
}
if root.Val < low {
return trimBST(root.Right, low, high)
}
root.Left = trimBST(root.Left, low, high)
root.Right = trimBST(root.Right, low, high)
return root
}
``````

Apr 8, 2023